RD Sharma XII Vol 2 2018 Solutions for Class 12 Science Math Chapter 3 - Differential Equations

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RD Sharma XII Vol 2 2018 Solutions for Class 12 Science Math Chapter 3 Differential Equations are provided here with simple step-by-step explanations. These solutions for Differential Equations are extremely popular among class 12 Science students for Math Differential Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 2 2018 Book of class 12 Science Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 2 2018 Solutions. All RD Sharma XII Vol 2 2018 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 22.106:

Question 1:

$\frac{d y}{d x} + 2 y = e^{3 x}$

Answer:

$We have, \frac{d y}{d x} + 2 y = e^{3 x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 Q = e^{3 x} ∴ I . F . = e^{\int P d x} = e^{\int 2 d x} = e^{2 x} Multiplying both sides of (1) by e^{2 x}, we get e^{2 x} (\frac{d y}{d x} + 2 y) = e^{2 x} e^{3 x} \Rightarrow e^{2 x} \frac{d y}{d x} + 2 e^{2 x} y = e^{5 x} Integrating both sides with respect to x, we get y e^{2 x} = \int e^{5 x} d x + C \Rightarrow y e^{2 x} = \frac{e^{5 x}}{5} + C \Rightarrow y = \frac{1}{5} e^{3 x} + C e^{- 2 x} Hence, y = \frac{1}{5} e^{3 x} + C e^{- 2 x} is the required solution .$

Page No 22.106:

Question 2:

$4 \frac{d y}{d x} + 8 y = 5 e^{- 3 x}$

Answer:

We have,
$4 \frac{d y}{d x} + 8 y = 5 e^{- 3 x}$

$\Rightarrow \frac{d y}{d x} + 2 y = \frac{5}{4} e^{- 3 x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 Q = \frac{5}{4} e^{- 3 x} ∴ I . F . = e^{\int P d x} = e^{\int 2 d x} = e^{2 x} Multiplying both sides of (1) by e^{2 x}, we get e^{2 x} (\frac{d y}{d x} + 2 y) = \frac{5}{4} e^{2 x} e^{- 3 x} \Rightarrow e^{2 x} \frac{d y}{d x} + 2 e^{2 x} y = \frac{5}{4} e^{- x} Integrating both sides with respect to x, we get y e^{2 x} = \frac{5}{4} \int e^{- x} d x + C \Rightarrow y e^{2 x} = - \frac{5}{4} e^{- x} + C \Rightarrow y = \frac{5}{4} e^{- 3 x} + C e^{- 2 x} Hence, y = \frac{5}{4} e^{- 3 x} + C e^{- 2 x} is the required solution .$

Page No 22.106:

Question 3:

$\frac{d y}{d x} + 2 y = 6 e^{x}$

Answer:

$We have, \frac{d y}{d x} + 2 y = 6 e^{x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 Q = 6 e^{x} ∴ I . F . = e^{\int P d x} = e^{\int 2 d x} = e^{2 x} Multiplying both sides of (1) by e^{2 x}, we get e^{2 x} (\frac{d y}{d x} + 2 y) = 6 e^{2 x} e^{x} \Rightarrow e^{2 x} \frac{d y}{d x} + 2 e^{2 x} y = 6 e^{3 x} Integrating both sides with respect to x, we get y e^{2 x} = 6 \int e^{3 x} d x + C \Rightarrow y e^{2 x} = 6 \frac{e^{3 x}}{3} + C \Rightarrow y e^{2 x} = 2 e^{3 x} + C Hence, y e^{2 x} = 2 e^{3 x} + C is the required solution .$

Page No 22.106:

Question 4:

$\frac{d y}{d x} + y = e^{- 2 x}$

Answer:

$We have, \frac{d y}{d x} + y = e^{- 2 x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 1 Q = e^{- 2 x} ∴ I . F . = e^{\int P d x} = e^{\int 1 d x} = e^{x} Multiplying both sides of (1) by e^{x}, we get e^{x} (\frac{d y}{d x} + y) = e^{x} e^{- 2 x} \Rightarrow e^{x} \frac{d y}{d x} + e^{x} y = e^{- x} Integrating both sides with respect to x, we get y e^{x} = \int e^{- x} d x + C \Rightarrow y e^{x} = - e^{- x} + C \Rightarrow y = - e^{- 2 x} + C e^{- x} Hence, y = - e^{- 2 x} + C e^{- x} is the required solution .$

Page No 22.106:

Question 5:

$x \frac{d y}{d x} = x + y$

Answer:

$We have, x \frac{d y}{d x} = x + y \Rightarrow \frac{d y}{d x} = 1 + \frac{1}{x} y \Rightarrow \frac{d y}{d x} - \frac{1}{x} y = 1 . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{1}{x} Q = 1 ∴ I . F . = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = e^{\log \frac{1}{x}} = \frac{1}{x} Multiplying both sides of (1) by \frac{1}{x}, we get \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = \frac{1}{x} \times 1 \Rightarrow \frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2}} y = \frac{1}{x} Integrating both sides with respect to x, we get y \frac{1}{x} = \int \frac{1}{x} d x + C \Rightarrow \frac{y}{x} = \log |x| + C Hence, \frac{y}{x} = \log |x| + C is the required solution .$

Page No 22.106:

Question 6:

$\frac{d y}{d x} + 2 y = 4 x$

Answer:

$We have, \frac{d y}{d x} + 2 y = 4 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 Q = 4 x ∴ I . F . = e^{\int P d x} = e^{\int 2 d x} = e^{2 x} Multiplying both sides of (1) by e^{2 x}, we get e^{2 x} (\frac{d y}{d x} + 2 y) = e^{2 x} 4 x \Rightarrow e^{2 x} \frac{d y}{d x} + 2 e^{2 x} y = e^{2 x} 4 x Integrating both sides with respect to x, we get y e^{2 x} = 4 \int x e^{2 x} d x + C \Rightarrow y e^{2 x} = 4 \int \underset{I}{x} \underset{II}{e^{2 x}} d x + C \Rightarrow y e^{2 x} = 4 x \int e^{2 x} d x - 4 \int [\frac{d}{d x} (x) \int e^{2 x} d x] d x + C \Rightarrow y e^{2 x} = 4 x \frac{e^{2 x}}{2} - 4 \times \frac{1}{2} \int e^{2 x} d x + C \Rightarrow y e^{2 x} = 2 x e^{2 x} - 4 \times \frac{1}{4} e^{2 x} + C \Rightarrow y e^{2 x} = 2 x e^{2 x} - e^{2 x} + C \Rightarrow y e^{2 x} = (2 x - 1) e^{2 x} + C \Rightarrow y = (2 x - 1) + C e^{- 2 x} Hence, y = (2 x - 1) + C e^{- 2 x} is the required solution .$

Page No 22.106:

Question 7:

$x \frac{d y}{d x} + y = x e^{x}$

Answer:

$We have, x \frac{d y}{d x} + y = x e^{x} \Rightarrow \frac{d y}{d x} + \frac{1}{x} y = e^{x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \frac{1}{x} Q = e^{x} ∴ I . F . = e^{\int P d x} = e^{\int \frac{1}{x} d x} = e^{\log x} = x Multiplying both sides of (1) by x, we get x (\frac{d y}{d x} + \frac{1}{x} y) = x e^{x} \Rightarrow x \frac{d y}{d x} + y = x e^{x} Integrating both sides with respect to x, we get x y = \int x e^{x} d x + C \Rightarrow x y = \int \underset{I}{x} \underset{II}{e^{x}} d x + C \Rightarrow x y = x \int e^{x} d x - \int (\frac{d}{d x} (x) \int e^{x} d x) d x + C \Rightarrow x y = x e^{x} - e^{x} + C \Rightarrow x y = (x - 1) e^{x} + C \Rightarrow y = (\frac{x - 1}{x}) e^{x} + \frac{C}{x} Hence, y = (\frac{x - 1}{x}) e^{x} + \frac{C}{x} is the required solution .$

Page No 22.106:

Question 8:

$\frac{d y}{d x} + \frac{4 x}{x^{2} + 1} y + \frac{1}{{(x^{2} + 1)}^{2}} = 0$

Answer:

$We have, \frac{d y}{d x} + \frac{4 x}{x^{2} + 1} y + \frac{1}{{(x^{2} + 1)}^{2}} = 0 \Rightarrow \frac{d y}{d x} + \frac{4 x}{x^{2} + 1} y = - \frac{1}{{(x^{2} + 1)}^{2}} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \frac{4 x}{x^{2} + 1} Q = - \frac{1}{{(x^{2} + 1)}^{2}} ∴ I . F . = e^{\int P d x} = e^{2 \int \frac{2 x}{x^{2} + 1} d x} = e^{2 \log |x^{2} + 1|} = {(x^{2} + 1)}^{2} Multiplying both sides of (1) by {(x^{2} + 1)}^{2}, we get {(x^{2} + 1)}^{2} (\frac{d y}{d x} + \frac{4 x}{x^{2} + 1} y) = {(x^{2} + 1)}^{2} [- \frac{1}{{(x^{2} + 1)}^{2}}] \Rightarrow {(x^{2} + 1)}^{2} \frac{d y}{d x} + 4 x (x^{2} + 1) y = - 1 Integrating both sides with respect to x, we get {(x^{2} + 1)}^{2} y = - \int d x + C \Rightarrow {(x^{2} + 1)}^{2} y = - x + C Hence, {(x^{2} + 1)}^{2} y = - x + C is the required solution .$

Page No 22.106:

Question 9:

$x \frac{d y}{d x} + y = x \log x$

Answer:

We have,
$x \frac{d y}{d x} + y = x \log x$
Dividing both sides by x, we get
$\frac{d y}{d x} + \frac{y}{x} = \log x Comparing with \frac{d y}{d x} + P y = Q, we get P = \frac{1}{x} Q = \log x Now, I . F . = e^{\int P d x} = e^{\int \frac{1}{x} d x} = e^{\log |x|} = x So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow x y = \int \underset{II}{x} \underset{I}{\log x} d x + C \Rightarrow x y = \log x \int x d x - \int [\frac{d}{d x} (\log x) \int x d x] d x + C \Rightarrow x y = \frac{x^{2} \log x}{2} - \int \frac{x}{2} d x + C \Rightarrow x y = \frac{x^{2} \log x}{2} - \frac{x^{2}}{4} + C \Rightarrow 4 x y = 2 x^{2} \log x - x^{2} + K (where, K = 2 C)$

Page No 22.106:

Question 10:

$x \frac{d y}{d x} - y = (x - 1) e^{x}$

Answer:

$We have, x \frac{d y}{d x} - y = (x - 1) e^{x} \Rightarrow \frac{d y}{d x} - \frac{1}{x} y = (\frac{x - 1}{x}) e^{x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{1}{x} Q = (\frac{x - 1}{x}) e^{x} ∴ I . F . = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x} Multiplying both sides of (1) by I . F . = \frac{1}{x}, we get \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = \frac{1}{x} (\frac{x - 1}{x}) e^{x} \Rightarrow \frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2}} y = (\frac{x - 1}{x^{2}}) e^{x} Integrating both sides with respect to x, we get \frac{1}{x} y = \int (\frac{1}{x} - \frac{1}{x^{2}}) e^{x} d x + C \Rightarrow \frac{1}{x} y = \frac{e^{x}}{x} + C \Rightarrow y = e^{x} + C x Hence, y = e^{x} + C x is the required solution .$

Page No 22.106:

Question 11:

$\frac{d y}{d x} + \frac{y}{x} = x^{3}$

Answer:

$We have, \frac{d y}{d x} + \frac{y}{x} = x^{3} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \frac{1}{x} Q = x^{3} ∴ I . F . = e^{\int P d x} = e^{\int \frac{1}{x} d x} = e^{\log |x|} = x Multiplying both sides of (1) by x, we get x (\frac{d y}{d x} + \frac{1}{x} y) = x x^{3} \Rightarrow x \frac{d y}{d x} + y = x^{4} Integrating both sides with respect to x, we get x y = \int x^{4} d x + C \Rightarrow x y = \frac{x^{5}}{5} + C \Rightarrow 5 x y = x^{5} + 5 C \Rightarrow 5 x y = x^{5} + K (where, K = 5 C) Hence, 5 x y = x^{5} + K is the required solution .$

Page No 22.106:

Question 12:

$\frac{d y}{d x} + y = \sin x$

Answer:

$We have, \frac{d y}{d x} + y = \sin x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 1 Q = \sin x ∴ I . F . = e^{\int P d x} = e^{\int d x} = e^{x} Multiplying both sides of (1) by e^{x}, we get e^{x} (\frac{d y}{d x} + y) = e^{x} \sin x \Rightarrow e^{x} \frac{d y}{d x} + e^{x} y = e^{x} \sin x Integrating both sides with respect to x, we get y e^{x} = \int e^{x} \sin x d x + C \Rightarrow y e^{x} = \frac{e^{x}}{2} (\sin x - \cos x) + C \Rightarrow y = C e^{- x} + \frac{1}{2} (\sin x - \cos x) Hence, y = C e^{- x} + \frac{1}{2} (\sin x - \cos x) is the required solution .$

Page No 22.106:

Question 13:

$\frac{d y}{d x} + y = \cos x$

Answer:

$We have, \frac{d y}{d x} + y = \cos x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 1 Q = \cos x ∴ I . F . = e^{\int P d x} = e^{\int d x} = e^{x} Multiplying both sides of (1) by e^{x}, we get e^{x} (\frac{d y}{d x} + y) = e^{x} \cos x \Rightarrow e^{x} \frac{d y}{d x} + e^{x} y = e^{x} \cos x Integrating both sides with respect to x, we get y e^{x} = \int e^{x} \cos x d x + C \Rightarrow y e^{x} = \frac{1}{2} \int e^{x} [(\cos x + \sin x) + (- \sin x + \cos x)] d x + C \Rightarrow y e^{x} = \frac{e^{x}}{2} (\cos x + \sin x) + C \Rightarrow y = \frac{1}{2} (\cos x + \sin x) + C e^{- x} Hence, y = \frac{1}{2} (\cos x + \sin x) + C e^{- x} is the required solution .$

Page No 22.106:

Question 14:

$\frac{d y}{d x} + 2 y = \sin x$

Answer:

$We have, \frac{d y}{d x} + 2 y = \sin x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 and Q = \sin x ∴ I . F . = e^{\int P d x} = e^{\int 2 d x} = e^{2 x} Multiplying both sides of (1) by I . F . = e^{2 x}, we get e^{2 x} (\frac{d y}{d x} + 2 y) = e^{2 x} \sin x \Rightarrow e^{2 x} \frac{d y}{d x} + 2 e^{2 x} y = e^{2 x} \sin x Integrating both sides with respect to x, we get y e^{2 x} = \int e^{2 x} \sin x d x + C \Rightarrow y e^{2 x} = \frac{1}{5} \int [2 e^{2 x} (2 \sin x - \cos x) + e^{2 x} (2 \cos x + \sin x)] d x + C Putting e^{2 x} (2 \sin x - \cos x) = t \Rightarrow [2 e^{2 x} (2 \sin x - \cos x) + e^{2 x} (2 \cos x + \sin x)] d x = d t ∴ y e^{2 x} = \frac{1}{5} \int d t + C \Rightarrow y e^{2 x} = \frac{t}{5} + C y e^{2 x} = \frac{e^{2 x}}{5} (2 \sin x - \cos x) + C \Rightarrow y = \frac{1}{5} (2 \sin x - \cos x) + C e^{- 2 x} Hence, y = \frac{1}{5} (2 \sin x - \cos x) + C e^{- 2 x} is the required solution .$

Page No 22.106:

Question 15:

$\frac{d y}{d x}$ = y tan x − 2 sin x

Answer:

$We have, \frac{d y}{d x} = y \tan x - 2 \sin x \frac{d y}{d x} - y \tan x = - 2 \sin x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \tan x Q = - 2 \sin x ∴ I . F . = e^{\int P d x} = e^{- \int \tan x d x} = e^{- \log |\sec x|} = \cos x Multiplying both sides of (1) by \cos x, we get \cos x (\frac{d y}{d x} - y \tan x) = - 2 \sin x \times \cos x \Rightarrow \cos x \frac{d y}{d x} - y \sin x = - \sin 2 x Integrating both sides with respect to x, we get y \cos x = - \int \sin 2 x d x + C \Rightarrow y \cos x = \frac{\cos 2 x}{2} + C \Rightarrow 2 y \cos x = \cos 2 x + 2 C \Rightarrow 2 y \cos x = \cos 2 x + K, (where k = 2 C) Hence, 2 y \cos x = \cos 2 x + K is the required solution .$

Page No 22.106:

Question 16:

$(1 + x^{2}) \frac{d y}{d x} + y = \tan^{- 1} x$

Answer:

$We have, (1 + x^{2}) \frac{d y}{d x} + y = \tan^{- 1} x \Rightarrow \frac{d y}{d x} + \frac{y}{1 + x^{2}} = \frac{\tan^{- 1} x}{1 + x^{2}} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \frac{1}{1 + x^{2}} Q = \frac{\tan^{- 1} x}{1 + x^{2}} ∴ I . F . = e^{\int P d x} = e^{\int \frac{1}{1 + x^{2}} d x} = e^{\tan^{- 1} x} Multiplying both sides of (1) by e^{\tan^{- 1} x}, we get e^{\tan^{- 1} x} (\frac{d y}{d x} + \frac{y}{1 + x^{2}}) = e^{\tan^{- 1} x} \frac{\tan^{- 1} x}{1 + x^{2}} \Rightarrow e^{\tan^{- 1} x} \frac{d y}{d x} + e^{\tan^{- 1} x} \frac{y}{1 + x^{2}} = e^{\tan^{- 1} x} \frac{\tan^{- 1} x}{1 + x^{2}} Integrating both sides with respect to x, we get e^{\tan^{- 1} x} y = \int \frac{\tan^{- 1} x \times e^{\tan^{- 1} x}}{1 + x^{2}} d x + C \Rightarrow e^{\tan^{- 1} x} y = I + C . . . . . (2) Here, I = \int \frac{\tan^{- 1} x \times e^{\tan^{- 1} x}}{1 + x^{2}} d x Putting \tan^{- 1} x = t, we get \Rightarrow \frac{1}{1 + x^{2}} d x = d t ∴ I = \int t e^{t} d t = t \int e^{t} d t - \int [\frac{d}{d t} (t) \int e^{t} d t] d t = t e^{t} - e^{t} = (t - 1) e^{t} = (\tan^{- 1} x - 1) e^{\tan^{- 1} x} Substituting the value of I in (2), we get e^{\tan^{- 1} x} y = (\tan^{- 1} x - 1) e^{\tan^{- 1} x} + C \Rightarrow y = \tan^{- 1} x - 1 + C e^{- \tan^{- 1} x} Hence, y = \tan^{- 1} x - 1 + C e^{- \tan^{- 1} x} is the required solution .$

Page No 22.106:

Question 17:

$\frac{d y}{d x}$ + y tan x = cos x

Answer:

$We have, \frac{d y}{d x} + y \tan x = \cos x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \tan x Q = \cos x ∴ I . F . = e^{\int P d x} = e^{\int \tan x d x} = e^{\log |\sec x|} = \sec x Multiplying both sides of (1) by \sec x, we get \sec x (\frac{d y}{d x} + y \tan x) = \cos x \times \sec x \Rightarrow \sec x \frac{d y}{d x} + y \sec x \tan x = 1 Integrating both sides with respect to x, we get y \sec x = \int d x + C \Rightarrow y \sec x = x + C Hence, y \sec x = x + C is the required solution .$

Page No 22.106:

Question 18:

$\frac{d y}{d x}$ + y cot x = x² cot x + 2x

Answer:

$We have, \frac{d y}{d x} + y \cot x = x^{2} \cot x + 2 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cot x Q = x^{2} \cot x + 2 x ∴ I . F . = e^{\int P d x} = e^{\int \cot x d x} = e^{\log |\sin x|} = \sin x Multiplying both sides of (1) by \sin x, we get \sin x (\frac{d y}{d x} + y \cot x) = \sin x (x^{2} \cot x + 2 x) \Rightarrow \sin x \frac{d y}{d x} + y \cos x = x^{2} \cos x + 2 x \sin x Integrating both sides with respect to x, we get y \sin x = \int \underset{I}{x^{2}} \underset{II}{\cos x} d x + \int 2 x \sin x d x + C \Rightarrow y \sin x = x^{2} \int \cos x d x - \int [\frac{d}{d x} (x^{2}) \int \cos x d x] d x + \int 2 x \sin x d x + C \Rightarrow y \sin x = x^{2} \sin x - \int 2 x \sin x d x + \int 2 x \sin x d x + C \Rightarrow y \sin x = x^{2} \sin x + C Hence, y \sin x = x^{2} \sin x + C is the required solution .$

Page No 22.106:

Question 19:

$\frac{d y}{d x} + y \tan x = x^{2} \cos^{2} x$

Answer:

We have,
$\frac{d y}{d x} + y \tan x = x^{2} \cos^{2} x$
$Comparing with \frac{d y}{d x} + P y = Q, we get P = \tan x Q = x^{2} \cos^{2} x Now, I . F . = e^{\int \tan x d x} = e^{\log |\sec x|} = \sec x Therefore, solution is given by y \times I . F . = \int x^{2} \cos^{2} x \times I . F . d x + C \Rightarrow y \sec x = \int x^{2} \cos x d x + C \Rightarrow y \sec x = I + C Where, I = \int \underset{II}{x^{2}} \underset{I}{\cos x} d x + C \Rightarrow I = x^{2} \int \cos x d x - \int [\frac{d}{d x} (x^{2}) \int \cos x d x] d x \Rightarrow I = x^{2} \sin x - 2 \int x \sin x d x \Rightarrow I = x^{2} \sin x - 2 \int \underset{I}{x} \underset{II}{\sin x} d x \Rightarrow I = x^{2} \sin x - 2 x \int \sin x d x + 2 \int [\frac{d}{d x} (x) \int \sin x d x] d x \Rightarrow I = x^{2} \sin x + 2 x \cos x - 2 \int \cos x d x \Rightarrow I = x^{2} \sin x + 2 x \cos x - 2 \sin x \Rightarrow I = x^{2} \sin x + 2 x \cos x - 2 \sin x ∴ y \sec x = x^{2} \sin x + 2 x \cos x - 2 \sin x + C \Rightarrow y \sec x = x^{2} \sin x + 2 x \cos x - 2 \sin x + C$

Page No 22.106:

Question 20:

$(1 + x^{2}) \frac{d y}{d x} + y = e^{\tan^{- 1} x}$

Answer:

$We have, (1 + x^{2}) \frac{d y}{d x} + y = e^{\tan^{- 1} x} \Rightarrow \frac{d y}{d x} + \frac{y}{1 + x^{2}} = \frac{e^{\tan^{- 1} x}}{1 + x^{2}} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \frac{1}{1 + x^{2}} Q = \frac{e^{\tan^{- 1} x}}{1 + x^{2}} ∴ I . F . = e^{\int P d x} = e^{\int \frac{1}{1 + x^{2}} d x} = e^{\tan^{- 1} x} Multiplying both sides of (1) by e^{\tan^{- 1} x}, we get e^{\tan^{- 1} x} (\frac{d y}{d x} + \frac{y}{1 + x^{2}}) = e^{\tan^{- 1} x} \frac{e^{\tan^{- 1} x}}{1 + x^{2}} \Rightarrow e^{\tan^{- 1} x} \frac{d y}{d x} + \frac{y e^{\tan^{- 1} x}}{1 + x^{2}} = e^{\tan^{- 1} x} \frac{e^{\tan^{- 1} x}}{1 + x^{2}} Integrating both sides with respect to x, we get y e^{\tan^{- 1} x} = \int \frac{e^{2 \tan^{- 1} x}}{1 + x^{2}} d x + C \Rightarrow y e^{\tan^{- 1} x} = I + C . . . . . (2) Here, I = \int \frac{e^{2 \tan^{- 1} x}}{1 + x^{2}} d x Putting \tan^{- 1} x = t, we get \frac{1}{1 + x^{2}} d x = d t ∴ I = \int e^{2 t} d t = \frac{e^{2 t}}{2} = \frac{e^{2 \tan^{- 1} x}}{2} Putting the value of I in (2), we get y e^{\tan^{- 1} x} = \frac{e^{2 \tan^{- 1} x}}{2} + C \Rightarrow 2 y e^{\tan^{- 1} x} = e^{2 \tan^{- 1} x} + 2 C \Rightarrow 2 y e^{\tan^{- 1} x} = e^{2 \tan^{- 1} x} + k, (where k = 2 C) Hence, 2 y e^{\tan^{- 1} x} = e^{2 \tan^{- 1} x} + k is the required solution .$

Page No 22.106:

Question 21:

x dy = (2y + 2x⁴ + x²) dx

Answer:

$We have, x d y = (2 y + 2 x^{4} + x^{2}) d x \Rightarrow \frac{d y}{d x} = \frac{2}{x} y + 2 x^{3} + x \Rightarrow \frac{d y}{d x} - \frac{2}{x} y = 2 x^{3} + x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{2}{x} Q = 2 x^{3} + x ∴ I . F . = e^{\int P d x} = e^{- \int \frac{2}{x} d x} = e^{- 2 \log x} = \frac{1}{x^{2}} Multiplying both sides of (1) by \frac{1}{x^{2}}, we get \frac{1}{x^{2}} (\frac{d y}{d x} - \frac{2}{x} y) = \frac{1}{x^{2}} (2 x^{3} + x) \Rightarrow \frac{1}{x^{2}} \frac{d y}{d x} - \frac{2}{x^{3}} y = 2 x + \frac{1}{x} Integrating both sides with respect to x, we get \frac{1}{x^{2}} y = \int (2 x + \frac{1}{x}) d x + C \Rightarrow \frac{1}{x^{2}} y = x^{2} + \log x + C \Rightarrow y = x^{4} + x^{2} \log x + C x^{2} Hence, y = x^{4} + x^{2} \log x + C x^{2} is the required solution .$

Page No 22.106:

Question 22:

$(1 + y^{2}) + (x - e^{\tan^{- 1} y}) \frac{d y}{d x} = 0$

Answer:

$We have, (1 + y^{2}) + (x - e^{\tan^{- 1} y}) \frac{d y}{d x} = 0 \Rightarrow (x - e^{\tan^{- 1} y}) \frac{d y}{d x} = - (1 + y^{2}) \Rightarrow \frac{d y}{d x} = - \frac{(1 + y^{2})}{(x - e^{\tan^{- 1} y})} \Rightarrow \frac{d x}{d y} = - \frac{x - e^{\tan^{- 1} y}}{1 + y^{2}} \Rightarrow \frac{d x}{d y} + \frac{x}{1 + y^{2}} = \frac{e^{\tan^{- 1} y}}{1 + y^{2}} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = \frac{1}{1 + y^{2}} Q = \frac{e^{\tan^{- 1} y}}{1 + y^{2}} ∴ I . F . = e^{\int P d y} = e^{\int \frac{1}{1 + y^{2}} d y} = e^{\tan^{- 1} y} Multiplying both sides of (1) by e^{\tan^{- 1} y}, we get e^{\tan^{- 1} y} (\frac{d x}{d y} + \frac{x}{1 + y^{2}}) = e^{\tan^{- 1} y} \frac{e^{\tan^{- 1} y}}{1 + y^{2}} \Rightarrow e^{\tan^{- 1} y} \frac{d x}{d y} + \frac{x e^{\tan^{- 1} x}}{1 + y^{2}} = \frac{e^{2 \tan^{- 1} y}}{1 + y^{2}} Integrating both sides with respect to y, we get x e^{\tan^{- 1} y} = \int \frac{e^{2 \tan^{- 1} y}}{1 + y^{2}} d y + C \Rightarrow x e^{\tan^{- 1} y} = I + C . . . . . (2) Here, I = \int \frac{e^{2 \tan^{- 1} y}}{1 + y^{2}} d y Putting \tan^{- 1} y = t, we get \frac{1}{1 + y^{2}} d y = d t ∴ I = \int e^{2 t} d t = \frac{e^{2 t}}{2} = \frac{e^{2 \tan^{- 1} y}}{2} Putting the value of I in (2), we get x e^{\tan^{- 1} y} = \frac{e^{2 \tan^{- 1} y}}{2} + C \Rightarrow 2 x e^{\tan^{- 1} y} = e^{2 \tan^{- 1} y} + 2 C \Rightarrow 2 x e^{\tan^{- 1} y} = e^{2 \tan^{- 1} y} + k (where k = 2 C) Hence, 2 x e^{\tan^{- 1} y} = e^{2 \tan^{- 1} y} + k is the required solution .$

Page No 22.106:

Question 23:

$y^{2} \frac{d x}{d y} + x - \frac{1}{y} = 0$

Answer:

$We have, y^{2} \frac{d x}{d y} + x - \frac{1}{y} = 0 \Rightarrow y^{2} \frac{d x}{d y} + x = \frac{1}{y} \Rightarrow \frac{d x}{d y} + \frac{1}{y^{2}} x = \frac{1}{y^{3}} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = \frac{1}{y^{2}} Q = \frac{1}{y^{3}} ∴ I . F . = e^{\int P d y} = e^{\int \frac{1}{y^{2}} d y} = e^{\frac{- 1}{y}} Multiplying both sides of (1) by e^{\frac{- 1}{y}}, we get e^{\frac{- 1}{y}} (\frac{d x}{d y} + x \frac{1}{y^{2}}) = e^{\frac{- 1}{y}} \frac{1}{y^{3}} \Rightarrow e^{\frac{- 1}{y}} \frac{d x}{d y} + x \frac{1}{y^{2}} e^{\frac{- 1}{y}} = e^{\frac{- 1}{y}} \frac{1}{y^{3}} Integrating both sides with respect to y, we get x e^{\frac{- 1}{y}} = \int e^{\frac{- 1}{y}} \frac{1}{y^{3}} d y + C \Rightarrow x e^{\frac{- 1}{y}} = I + C . . . . . (2) where I = \int e^{\frac{- 1}{y}} \frac{1}{y^{3}} d y Putting t = \frac{1}{y}, we get d t = - \frac{1}{y^{2}} d y ∴ I = - \int \underset{I}{t} \underset{II}{e^{- t}} d t = - t \int e^{- t} d t + \int [\frac{d}{d t} (t) \int e^{- t} d t] d t = t e^{- t} + e^{- t} = (t + 1) e^{- t} = (\frac{1}{y} + 1) e^{- \frac{1}{y}} Putting the value of I in (2), we get x e^{\frac{- 1}{y}} = (\frac{1}{y} + 1) e^{- \frac{1}{y}} + C \Rightarrow x = (\frac{y + 1}{y}) + C e^{\frac{1}{y}} Hence, x = (\frac{y + 1}{y}) + C e^{\frac{1}{y}} is the required solution .$

Page No 22.106:

Question 24:

$(2 x - 10 y^{3}) \frac{d y}{d x} + y = 0$

Answer:

$We have, (2 x - 10 y^{3}) \frac{d y}{d x} + y = 0 \Rightarrow (2 x - 10 y^{3}) \frac{d y}{d x} = - y \Rightarrow \frac{d x}{d y} = - \frac{1}{y} (2 x - 10 y^{3}) \Rightarrow \frac{d x}{d y} + \frac{2}{y} x = 10 y^{2} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = \frac{2}{y} Q = 10 y^{2} ∴ I . F . = e^{\int P d y} = e^{\int \frac{2}{y} d y} = e^{2 \log y} = y^{2} Multiplying both sides of (1) by y^{2}, we get y^{2} (\frac{d x}{d y} + \frac{2}{y} x) = y^{2} \times 10 y^{2} \Rightarrow y^{2} \frac{d x}{d y} + \frac{2}{y} x y^{2} = 10 y^{4} Integrating both sides with respect to y, we get x y^{2} = \int 10 y^{4} d y + C \Rightarrow x y^{2} = 2 y^{5} + C \Rightarrow x = 2 y^{3} + C y^{- 2} Hence, x = 2 y^{3} + C y^{- 2} is the required solution .$

Page No 22.106:

Question 25:

(x + tan y) dy = sin 2y dx

Answer:

$We have, (x + \tan y) d y = \sin 2 y d x \Rightarrow \frac{d x}{d y} = x cosec 2 y + \frac{1}{2} \sec^{2} y \Rightarrow \frac{d x}{d y} - x cosec 2 y = \frac{1}{2} \sec^{2} y . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = - cosec 2 y Q = \frac{1}{2} \sec^{2} y ∴ I . F . = e^{\int P d y} = e^{- \int cosec 2 y d y} = e^{- \frac{1}{2} \log |\tan y|} = \frac{1}{\sqrt{\tan y}} Multiplying both sides of (1) by \frac{1}{\sqrt{\tan y}}, we get \frac{1}{\sqrt{\tan y}} (\frac{d x}{d y} - x cosec 2 y) = \frac{1}{2} \frac{1}{\sqrt{\tan y}} \times \sec^{2} y \Rightarrow \frac{1}{\sqrt{\tan y}} \frac{d x}{d y} - x cosec 2 y \frac{1}{\sqrt{\tan y}} = \frac{1}{2} \frac{1}{\sqrt{\tan y}} \times \sec^{2} y Integrating both sides with respect to y, we get \frac{1}{\sqrt{\tan y}} x = \int \frac{1}{2} \frac{1}{\sqrt{\tan y}} \times \sec^{2} y d y + C \Rightarrow \frac{x}{\sqrt{\tan y}} = I + C . . . . . (2) where I = \int \frac{1}{2} \frac{1}{\sqrt{\tan y}} \times \sec^{2} y d y Putting t = \tan y, we get d t = \sec^{2} y d y ∴ I = \frac{1}{2} \int \frac{1}{\sqrt{t}} \times d t = \sqrt{t} = \sqrt{\tan y} Putting the value of I in (2), we get \frac{x}{\sqrt{\tan y}} = \sqrt{\tan y} + C \Rightarrow x = \tan y + C \sqrt{\tan y} Hence, x = \tan y + C \sqrt{\tan y} is the required solution .$

Page No 22.106:

Question 26:

dx + xdy = e^−y sec²y dy

Answer:

$We have, d x + x d y = e^{- y} \sec^{2} y d y \Rightarrow d x = e^{- y} \sec^{2} y d y - x d y \Rightarrow \frac{d x}{d y} = e^{- y} \sec^{2} y - x \Rightarrow \frac{d x}{d y} + x = e^{- y} \sec^{2} y . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = 1 Q = e^{- y} \sec^{2} y ∴ I . F . = e^{\int P d y} = e^{\int d y} = e^{y} Multiplying both sides of (1) by e^{y}, we get e^{y} (\frac{d x}{d y} + x) = e^{y} e^{- y} \sec^{2} y \Rightarrow e^{y} \frac{d x}{d y} + x e^{y} = \sec^{2} y Integrating both sides with respect to y, we get x e^{y} = \int \sec^{2} y d y + C \Rightarrow x e^{y} = \tan y + C Hence, x e^{y} = \tan y + C is the required solution .$

Page No 22.106:

Question 27:

$\frac{d y}{d x}$ = y tan x − 2 sin x

Answer:

$We have, \frac{d y}{d x} = y \tan x - 2 \sin x \frac{d y}{d x} - y \tan x = - 2 \sin x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \tan x Q = - 2 \sin x ∴ I . F . = e^{\int P d x} = e^{- \int \tan x d x} = e^{- \log |\sec x|} = \cos x Multiplying both sides of (1) by \cos x, we get \cos x (\frac{d y}{d x} - y \tan x) = - 2 \sin x \times \cos x \Rightarrow \cos x \frac{d y}{d x} + y \sin x = - \sin 2 x Integrating both sides with respect to x, we get y \cos x = - \int \sin 2 x d x + C \Rightarrow y \cos x = \frac{\cos 2 x}{2} + C \Rightarrow y \cos x = \frac{1 - 2 \sin^{2} x}{2} + C \Rightarrow y \cos x = - \sin^{2} x + \frac{1}{2} + C \Rightarrow y \cos x = - \sin^{2} x + K (where k = \frac{1}{2} + C) \Rightarrow y = \sec x (- \sin^{2} x + K) Hence, y = \sec x (- \sin^{2} x + K) is the required solution .$

Page No 22.106:

Question 28:

$\frac{d y}{d x}$ + y cos x = sin x cos x

Answer:

$We have, \frac{d y}{d x} + y \cos x = \sin x \cos x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cos x Q = \sin x \cos x ∴ I . F . = e^{\int P d x} = e^{\int \cos x d x} = e^{\sin x} Multiplying both sides of (1) by e^{\sin x}, we get e^{\sin x} (\frac{d y}{d x} + y \cos x) = e^{\sin x} \sin x \cos x \Rightarrow e^{\sin x} \frac{d y}{d x} + e^{\sin x} y \cos x = e^{\sin x} \sin x \cos x Integrating both sides with respect to x, we get y e^{\sin x} = \int e^{\sin x} \sin x \cos x d x + C \Rightarrow y e^{\sin x} = I + C . . . . . (2) where I = \int e^{\sin x} \sin x \cos x d x Putting t = \sin x, we get d t = \cos x d x ∴ I = \int \underset{II}{e^{t}} \underset{I}{t} d t = t \int e^{t} d t - \int [\frac{d}{d t} (t) \int e^{t} d t] d t = t e^{t} - e^{t} = e^{t} (t - 1) = e^{\sin x} (\sin x - 1) Putting the value of I in (2), we get y e^{\sin x} = e^{\sin x} (\sin x - 1) + C \Rightarrow y = \sin x - 1 + C e^{- \sin x} Hence, y = \sin x - 1 + C e^{- \sin x} is the required solution .$

Page No 22.106:

Question 29:

Solve the following differential equations:

$(1 + x^{2}) \frac{d y}{d x} - 2 x y = (x^{2} + 2) (x^{2} + 1)$ [CBSE 2005]

Answer:

Given, $(1 + x^{2}) \frac{d y}{d x} - 2 x y = (x^{2} + 2) (x^{2} + 1)$
$\Rightarrow \frac{d y}{d x} - \frac{2 x}{(1 + x^{2})} y = (x^{2} + 2)$
This is a linear differential equation.
I.F. $= e^{\int - \frac{2 x}{1 + x^{2}} d x} = \frac{1}{1 + x^{2}}$
$y (\frac{1}{1 + x^{2}}) = \int (\frac{x^{2} + 2}{x^{2} + 1}) d x \Rightarrow y (\frac{1}{1 + x^{2}}) = \int (1 + \frac{1}{1 + x^{2}}) d x \Rightarrow y (\frac{1}{1 + x^{2}}) = x + \tan^{- 1} x + C \Rightarrow y = (x + \tan^{- 1} x + C) (1 + x^{2})$

Page No 22.106:

Question 30:

$(\sin x) \frac{d y}{d x} + y \cos x = 2 \sin^{2} x \cos x$

Answer:

$We have, (\sin x) \frac{d y}{d x} + y \cos x = 2 \sin^{2} x \cos x \Rightarrow \frac{d y}{d x} + y \cot x = 2 \sin x \cos x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cot x Q = 2 \sin x \cos x ∴ I . F . = e^{\int P d x} = e^{\int \cot x d x} = e^{\log |\sin x|} = \sin x Multiplying both sides of (1) by \sin x, we get \sin x (\frac{d y}{d x} + y \cot x) = \sin x \times 2 \sin x \cos x \Rightarrow \sin x \frac{d y}{d x} + y \cos x = 2 \sin^{2} x \cos x Integrating both sides with respect to x, we get y \sin x = 2 \int \sin^{2} x \cos x d x + C . . . . . (2) Putting \sin x = t \Rightarrow \cos x d x = d t Therefore, (2) becomes y \sin x = 2 \int t^{2} d t + C \Rightarrow y \sin x = \frac{2}{3} t^{3} + C \Rightarrow y \sin x = \frac{2}{3} \sin^{3} x + C Hence, y \sin x = \frac{2}{3} \sin^{3} x + C is the required solution .$

Page No 22.106:

Question 31:

$(x^{2} - 1) \frac{d y}{d x} + 2 (x + 2) y = 2 (x + 1)$

Answer:

$We have, (x^{2} - 1) \frac{d y}{d x} + 2 (x + 2) y = 2 (x + 1) \Rightarrow \frac{d y}{d x} + \frac{2 (x + 2)}{x^{2} - 1} y = \frac{2 (x + 1)}{x^{2} - 1} \Rightarrow \frac{d y}{d x} + \frac{2 (x + 2)}{x^{2} - 1} y = \frac{2}{x - 1} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \frac{2 (x + 2)}{x^{2} - 1} Q = \frac{2}{x - 1} ∴ I . F . = e^{\int P d x} = e^{\int \frac{2 (x + 2)}{x^{2} - 1} d x} = e^{\int \frac{2 x}{x^{2} - 1} + 4 \int \frac{1}{x^{2} - 1} d x} = e^{\log |x^{2} - 1| + 4 \times \frac{1}{2} \log |\frac{x - 1}{x + 1}|} = e^{\log |(x^{2} - 1) \times \frac{{(x - 1)}^{2}}{{(x + 1)}^{2}}|} = e^{\log |\frac{{(x - 1)}^{3}}{(x + 1)}|} = \frac{{(x - 1)}^{3}}{(x + 1)} Multiplying both sides of (1) by \frac{{(x - 1)}^{3}}{(x + 1)}, we get \frac{{(x - 1)}^{3}}{(x + 1)} (\frac{d y}{d x} + \frac{2 (x + 2)}{x^{2} - 1} y) = \frac{{(x - 1)}^{3}}{(x + 1)} \times \frac{2}{x - 1} \Rightarrow \frac{{(x - 1)}^{3}}{(x + 1)} \frac{d y}{d x} - \frac{2 (x + 2) {(x - 1)}^{2}}{{(x + 1)}^{2}} y = \frac{2 {(x - 1)}^{2}}{(x + 1)} Integrating both sides with respect to x, we get \frac{{(x - 1)}^{3}}{(x + 1)} y = \int \frac{2 {(x - 1)}^{2}}{(x + 1)} d x + C \Rightarrow \frac{{(x - 1)}^{3}}{(x + 1)} y = \int 2 \{\frac{{(x + 1)}^{2} - 4 x}{(x + 1)}\} d x + C \Rightarrow \frac{{(x - 1)}^{3}}{(x + 1)} y = \int 2 \{(x + 1) - \frac{4 x}{(x + 1)}\} d x + C \Rightarrow \frac{{(x - 1)}^{3}}{(x + 1)} y = \int 2 \{(x + 1) - \frac{4 (x + 1 - 1)}{(x + 1)}\} d x + C \Rightarrow \frac{{(x - 1)}^{3}}{(x + 1)} y = \int 2 \{(x + 1) - 4 + \frac{4}{(x + 1)}\} d x + C \Rightarrow \frac{{(x - 1)}^{3}}{(x + 1)} y = \int (2 x - 6 + \frac{8}{x + 1}) d x + C \Rightarrow \frac{{(x - 1)}^{3}}{(x + 1)} y = x^{2} - 6 x + 8 \log |x + 1| + C \Rightarrow y = \frac{(x + 1)}{{(x - 1)}^{3}} (x^{2} - 6 x + 8 \log |x + 1| + C) Hence, y = \frac{(x + 1)}{{(x - 1)}^{3}} (x^{2} - 6 x + 8 \log |x + 1| + C) is the required solution .$

Page No 22.106:

Question 32:

$x \frac{d y}{d x} + 2 y = x \cos x$

Answer:

$We have, x \frac{d y}{d x} + 2 y = x \cos x \Rightarrow \frac{d y}{d x} + \frac{2}{x} y = \cos x Comparing with \frac{d y}{d x} + P y = Q, we get P = \frac{2}{x} Q = \cos x Now, I . F . = e^{\int \frac{2}{x} d x} = e^{2 \log |x|} = x^{2} Solution is given by, y \times I . F . = \int \cos x \times I . F . d x + C \Rightarrow y x^{2} = \int x^{2} \cos x d x + C \Rightarrow x^{2} y = I + C . . . . . (1) Where, I = \int \underset{II}{x^{2}} \underset{I}{\cos x} d x + C \Rightarrow I = x^{2} \int \cos x d x - \int [\frac{d}{d x} (x^{2}) \int \cos x d x] d x \Rightarrow I = x^{2} \sin x - 2 \int x \sin x d x \Rightarrow I = x^{2} \sin x - 2 \int \underset{I}{x} \underset{II}{\sin x} d x \Rightarrow I = x^{2} \sin x - 2 x \int \sin x d x + 2 \int [\frac{d}{d x} (x) \int \sin x d x] d x \Rightarrow I = x^{2} \sin x + 2 x \cos x - 2 \int \cos x d x \Rightarrow I = x^{2} \sin x + 2 x \cos x - 2 \sin x \Rightarrow I = x^{2} \sin x + 2 x \cos x - 2 \sin x Therefore (1) becomes ∴ x^{2} y = x^{2} \sin x + 2 x \cos x - 2 \sin x + C$

Page No 22.106:

Question 33:

$\frac{d y}{d x} - y = x e^{x}$

Answer:

$We have, \frac{d y}{d x} - y = x e^{x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - 1 Q = e^{x} ∴ I . F . = e^{\int P d x} = e^{- \int d x} = e^{- x} Multiplying both sides of (1) by e^{- x}, we get e^{- x} (\frac{d y}{d x} - y) = x e^{x} e^{- x} \Rightarrow e^{- x} \frac{d y}{d x} - e^{- x} y = x Integrating both sides with respect to x, we get e^{- x} y = \int x d x + C \Rightarrow e^{- x} y = \frac{x^{2}}{2} + C \Rightarrow y = (\frac{x^{2}}{2} + C) e^{x} Hence, y = (\frac{x^{2}}{2} + C) e^{x} is the required solution .$

Page No 22.106:

Question 34:

$\frac{d y}{d x} + 2 y = x e^{4 x}$

Answer:

$We have, \frac{d y}{d x} + 2 y = x e^{4 x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 Q = x e^{4 x} ∴ I . F . = e^{\int P d x} = e^{\int 2 d x} = e^{2 x} Multiplying both sides of (1) by e^{2 x}, we get e^{2 x} (\frac{d y}{d x} + 2 y) = e^{2 x} \times x e^{4 x} \Rightarrow e^{2 x} \frac{d y}{d x} + 2 e^{2 x} y = x e^{6 x} Integrating both sides with respect to x, we get e^{2 x} y = \int \underset{II}{e^{6 x}} \underset{I}{x} d x + C \Rightarrow e^{2 x} y = x \int e^{6 x} d x - \int [\frac{d}{d x} (x) \int e^{6 x} d x] d x + C \Rightarrow e^{2 x} y = \frac{x e^{6 x}}{6} - \frac{e^{6 x}}{36} + C \Rightarrow y = \frac{x e^{4 x}}{6} - \frac{e^{4 x}}{36} + C e^{- 2 x} Hence, y = \frac{x e^{4 x}}{6} - \frac{e^{4 x}}{36} + C e^{- 2 x} is the required solution .$

Page No 22.106:

Question 35:

Solve the differential equation $(x + 2 y^{2}) \frac{d y}{d x} = y$ , given that when x = 2, y = 1.

Answer:

$We have, (x + 2 y^{2}) \frac{d y}{d x} = y \Rightarrow \frac{d x}{d y} = \frac{1}{y} (x + 2 y^{2}) \Rightarrow \frac{d x}{d y} - \frac{1}{y} x = 2 y . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = - \frac{1}{y} Q = 2 y ∴ I . F . = e^{\int P d y} = e^{- \int \frac{1}{y} d y} = e^{- \log y} = \frac{1}{y} Multiplying both sides of (1) by \frac{1}{y}, we get \frac{1}{y} (\frac{d x}{d y} - \frac{1}{y} x) = \frac{1}{y} \times 2 y \Rightarrow \frac{1}{y} \frac{d x}{d y} - \frac{1}{y^{2}} x = 2 Integrating both sides with respect to y, we get x \frac{1}{y} = \int 2 d y + C \Rightarrow x \frac{1}{y} = 2 y + C \Rightarrow x = 2 y^{2} + C y . . . . . (2) Now, y = 1 at x = 2 ∴ 2 = 2 + C \Rightarrow C = 0 Putting the value of C in (2), we get x = 2 y^{2} Hence, x = 2 y^{2} is the required solution .$

Page No 22.107:

Question 36:

Find one-parameter families of solution curves of the following differential equations:
(or Solve the following differential equations)
(i) $\frac{d y}{d x} + 3 y = e^{m x}$ , m is a given real number

(ii) $\frac{d y}{d x} - y = \cos 2 x$

(iii) $x \frac{d y}{d x} - y = (x + 1) e^{- x}$

(iv) $x \frac{d y}{d x} + y = x^{4}$

(v) $(x \log x) \frac{d y}{d x} + y = \log x$

(vi) $\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = x^{2} + 2$

(vii) $\frac{d y}{d x} + y \cos x = e^{\sin x} \cos x$

(viii) $(x + y) \frac{d y}{d x} = 1$

(ix) $\frac{d y}{d x} \cos^{2} x = \tan x - y$

(x) $e^{- y} \sec^{2} y d y = d x + x d y$

(xi) $x \log x \frac{d y}{d x} + y = 2 \log x$

(xii) $x \frac{d y}{d x} + 2 y = x^{2} \log x$

Answer:

$(i) We have, \frac{d y}{d x} + 3 y = e^{m x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 3 Q = e^{m x} ∴ I . F . = e^{\int P d x} = e^{\int 3 d x} = e^{3 x} Multiplying both sides of (1) by e^{3 x}, we get e^{3 x} (\frac{d y}{d x} + 3 y) = e^{3 x} e^{m x} \Rightarrow e^{3 x} \frac{d y}{d x} + 3 e^{3 x} y = e^{(m + 3) x} Integrating both sides with respect to x, we get y e^{3 x} = \int e^{(m + 3) x} d x + C (when m + 3 \neq 0) \Rightarrow y e^{3 x} = \frac{e^{(m + 3) x}}{m + 3} + C \Rightarrow y = \frac{e^{m x}}{m + 3} + C e^{- 3 x} y e^{3 x} = \int e^{0 \times x} d x + C (when m + 3 = 0) \Rightarrow y e^{3 x} = \int d x + C \Rightarrow y e^{3 x} = x + C \Rightarrow y = (x + C) e^{- 3 x} Hence, y = \frac{e^{m x}}{m + 3} + C e^{- 3 x}, where m + 3 \neq 0 and y = (x + C) e^{- 3 x}, where m + 3 = 0 are required solutions .$

$(ii) We have, \frac{d y}{d x} - y = \cos 2 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - 1 Q = \cos 2 x ∴ I . F . = e^{\int P d x} = e^{- \int d x} = e^{- x} Multiplying both sides of (1) by e^{- x}, we get e^{- x} (\frac{d y}{d x} - y) = e^{- x} \cos 2 x \Rightarrow e^{- x} \frac{d y}{d x} - e^{- x} y = e^{- x} \cos 2 x Integrating both sides with respect to x, we get y e^{- x} = \int e^{- x} \cos 2 x d x + C \Rightarrow y e^{- x} = I + C . . . . . (2) Where, I = \int e^{- x} \cos 2 x d x . . . . . (3) \Rightarrow I = \frac{1}{2} e^{- x} \sin 2 x - \frac{1}{2} \int (- e^{- x} \sin 2 x) d x \Rightarrow I = \frac{1}{2} e^{- x} \sin 2 x + \frac{1}{2} \int e^{- x} \sin 2 x d x \Rightarrow I = \frac{1}{2} e^{- x} \sin 2 x - \frac{1}{4} e^{- x} \cos 2 x - \frac{1}{2} \times \frac{1}{2} \int [(- e^{- x}) \times (- \cos 2 x)] d x \Rightarrow I = \frac{1}{2} e^{- x} \sin 2 x - \frac{1}{4} e^{- x} \cos 2 x - \frac{1}{4} \int e^{- x} \cos 2 x d x \Rightarrow I = \frac{1}{2} e^{- x} \sin 2 x - \frac{1}{4} e^{- x} \cos 2 x - \frac{1}{4} I [From (3)] \Rightarrow \frac{5}{4} I = \frac{1}{2} e^{- x} \sin 2 x - \frac{1}{4} e^{- x} \cos 2 x \Rightarrow 5 I = 2 e^{- x} \sin 2 x - e^{- x} \cos 2 x \Rightarrow I = \frac{e^{- x}}{5} (2 \sin 2 x - \cos 2 x) . . . . . (4) From (2) and (4) we get \Rightarrow y e^{- x} = \frac{e^{- x}}{5} (2 \sin 2 x - \cos 2 x) + C \Rightarrow y = \frac{1}{5} (2 \sin 2 x - \cos 2 x) + C e^{x} Hence, y = \frac{1}{5} (2 \sin 2 x - \cos 2 x) + C e^{x} is the required solution .$

$(iii) We have, x \frac{d y}{d x} - y = (x + 1) e^{- x} \Rightarrow \frac{d y}{d x} - \frac{1}{x} y = (\frac{x + 1}{x}) e^{- x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{1}{x} Q = (\frac{x + 1}{x}) e^{- x} ∴ I . F . = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x} Multiplying both sides of (1) by \frac{1}{x}, we get \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = \frac{1}{x} (\frac{x + 1}{x}) e^{- x} \Rightarrow \frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2}} y = (\frac{x + 1}{x^{2}}) e^{- x} Integrating both sides with respect to x, we get \frac{1}{x} y = \int (\frac{1}{x} + \frac{1}{x^{2}}) e^{- x} d x + C . . . . . (2) Putting \frac{1}{x} e^{- x} = t \Rightarrow (- \frac{1}{x} e^{- x} - \frac{1}{x^{2}} e^{- x}) d x = d t \Rightarrow (\frac{1}{x} + \frac{1}{x^{2}}) e^{- x} d x = - d t Therefore (2) becomes \frac{1}{x} y = - \int d t + C \Rightarrow \frac{1}{x} y = - t + C \Rightarrow \frac{1}{x} y = - \frac{1}{x} e^{- x} + C \Rightarrow y = - e^{- x} + C x Hence, y = - e^{- x} + C x is the required solution .$

$(iv) We have, x \frac{d y}{d x} + y = x^{4} \Rightarrow \frac{d y}{d x} + \frac{1}{x} y = x^{3} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \frac{1}{x} Q = x^{3} ∴ I . F . = e^{\int P d x} = e^{\int \frac{1}{x} d x} = e^{\log x} = x Multiplying both sides of (1) by x, we get x (\frac{d y}{d x} + \frac{1}{x} y) = x . x^{3} \Rightarrow x \frac{d y}{d x} + y = x^{4} Integrating both sides with respect to x, we get x y = \int x^{4} d x + C \Rightarrow x y = \frac{x^{5}}{5} + C \Rightarrow y = \frac{x^{4}}{5} + \frac{C}{x} Hence, y = \frac{x^{4}}{5} + \frac{C}{x} is the required solution .$

$(v) We have, (x \log x) \frac{d y}{d x} + y = \log x Dividing both sides by x \log x, we get \frac{d y}{d x} + \frac{y}{x \log x} = \frac{\log x}{x \log x} \Rightarrow \frac{d y}{d x} + \frac{y}{x \log x} = \frac{1}{x} \Rightarrow \frac{d y}{d x} + (\frac{1}{x \log x}) y = \frac{1}{x} Comparing with \frac{d y}{d x} + P y = Q, we get P = \frac{1}{x \log x} Q = \frac{1}{x} Now, I . F . = e^{\int P d x} = e^{\int \frac{1}{x \log x} d x} = e^{\log (\log x)} = \log x So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow y \log x = \int \frac{1}{x} \times \log x d x + C \Rightarrow y \log x = \frac{{(\log x)}^{2}}{2} + C \Rightarrow y = \frac{1}{2} \log x + \frac{C}{\log x}$

$(vi) We have, \frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = x^{2} + 2 . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{2 x}{1 + x^{2}} Q = x^{2} + 2 ∴ I . F . = e^{\int P d x} = e^{- \int \frac{2 x}{1 + x^{2}} d x} = e^{- \log |1 + x^{2}|} = \frac{1}{1 + x^{2}} Multiplying both sides of (1) by \frac{1}{1 + x^{2}}, we get \frac{1}{1 + x^{2}} (\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}}) = \frac{1}{1 + x^{2}} (x^{2} + 2) \Rightarrow \frac{1}{1 + x^{2}} \frac{d y}{d x} - \frac{2 x y}{{(1 + x^{2})}^{2}} = \frac{x^{2} + 2}{x^{2} + 1} Integrating both sides with respect to x, we get \frac{1}{1 + x^{2}} y = \int \frac{x^{2} + 2}{x^{2} + 1} d x + C \Rightarrow \frac{1}{1 + x^{2}} y = \int \frac{x^{2} + 1 + 1}{x^{2} + 1} d x + C \Rightarrow \frac{1}{1 + x^{2}} y = \int d x + \int \frac{1}{x^{2} + 1} d x + C \Rightarrow \frac{1}{1 + x^{2}} y = x + \tan^{- 1} x + C \Rightarrow y = (1 + x^{2}) (x + \tan^{- 1} x + C) Hence, y = (1 + x^{2}) (x + \tan^{- 1} x + C) is the required solution .$

$(vii) We have, \frac{d y}{d x} + y \cos x = e^{\sin x} \cos x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cos x Q = e^{\sin x} \cos x ∴ I . F . = e^{\int P d x} = e^{\int \cos x d x} = e^{\sin x} Multiplying both sides of (1) by e^{\sin x}, we get e^{\sin x} (\frac{d y}{d x} + y \cos x) = e^{\sin x} \times e^{\sin x} \cos x \Rightarrow e^{\sin x} \frac{d y}{d x} + y e^{\sin x} \cos x = e^{2 \sin x} \cos x Integrating both sides with respect to x, we get e^{\sin x} y = \int e^{2 \sin x} \cos x d x + C \Rightarrow e^{\sin x} y = I + C . . . . . (2) Where, I = \int e^{2 \sin x} \cos x d x Putting t = \sin x, we get d t = \cos x d x ∴ I = \int e^{2 t} d t = \frac{e^{2 t}}{2} = \frac{e^{2 \sin x}}{2} Putting the value of I in (2), we get e^{\sin x} y = \frac{e^{2 \sin x}}{2} + C \Rightarrow y = \frac{e^{\sin x}}{2} + C e^{- \sin x} Hence, y = \frac{e^{\sin x}}{2} + C e^{- \sin x} is the required solution .$

$(viii) We have, (x + y) \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{x + y} \Rightarrow \frac{d x}{d y} = x + y \Rightarrow \frac{d x}{d y} - x = y . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = - 1 Q = y ∴ I . F . = e^{\int P d y} = e^{\int - 1 d y} = e^{- y} Multiplying both sides of (1) by e^{- y}, we get e^{- y} (\frac{d x}{d y} - x) = e^{- y} y \Rightarrow e^{- y} \frac{d x}{d y} - e^{- y} x = e^{- y} y Integrating both sides with respect to y, we get e^{- y} x = \int \underset{I}{y} \underset{II}{e^{- y}} d y + C \Rightarrow e^{- y} x = y \int e^{- y} d y - \int [\frac{d}{d y} (y) \int e^{- y} d y] d y + C \Rightarrow e^{- y} x = - y e^{- y} - e^{- y} + C \Rightarrow e^{- y} x + y e^{- y} + e^{- y} = C \Rightarrow (x + y + 1) e^{- y} = C \Rightarrow (x + y + 1) = C e^{y} Hence, (x + y + 1) = C e^{y} is the required solution .$

$(ix) We have, \frac{d y}{d x} \cos^{2} x = \tan x - y \Rightarrow \frac{d y}{d x} + \frac{1}{\cos^{2} x} y = \tan x \sec^{2} x \Rightarrow \frac{d y}{d x} + y \sec^{2} x = \tan x \sec^{2} x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \sec^{2} x Q = \tan x \sec^{2} x ∴ I . F . = e^{\int P d x} = e^{\int \sec^{2} x d x} = e^{\tan x} Multiplying both sides of (1) by e^{\tan x}, we get e^{\tan x} (\frac{d y}{d x} + y \sec^{2} x) = e^{\tan x} \tan x \sec^{2} x \Rightarrow e^{\tan x} \frac{d y}{d x} + y e^{\tan x} \sec^{2} x = e^{\tan x} \tan x \sec^{2} x Integrating both sides with respect to x, we get e^{\tan x} y = \int e^{\tan x} \tan x \sec^{2} x d x + C \Rightarrow e^{\tan x} y = I + C . . . . . (2) Where, I = \int e^{\tan x} \tan x \sec^{2} x d x Putting t = \tan x, we get d t = \sec^{2} x d x ∴ I = \int \underset{I}{t} \underset{II}{e^{t}} d t = t \int e^{t} d t - \int [\frac{d}{d t} (t) \int e^{t} d t] d t = t e^{t} - e^{t} = (t - 1) e^{t} = (\tan x - 1) e^{\tan x} Putting the value of I in (2), we get e^{\tan x} y = (\tan x - 1) e^{\tan x} + C \Rightarrow y = (\tan x - 1) + C e^{- \tan x} Hence, y = (\tan x - 1) + C e^{- \tan x} is the required solution .$

$(x) We have, e^{- y} \sec^{2} y d y = d x + x d y \Rightarrow d x = e^{- y} \sec^{2} y d y - x d y \Rightarrow \frac{d x}{d y} = e^{- y} \sec^{2} y - x \Rightarrow \frac{d x}{d y} + x = e^{- y} \sec^{2} y . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = 1 Q = e^{- y} \sec^{2} y ∴ I . F . = e^{\int P d y} = e^{\int d y} = e^{y} Multiplying both sides of (1) by e^{y}, we get e^{y} (\frac{d x}{d y} + x) = e^{y} e^{- y} \sec^{2} y \Rightarrow e^{y} \frac{d x}{d y} + e^{y} x = \sec^{2} y Integrating both sides with respect to y, we get e^{y} x = \int \sec^{2} y d y + C \Rightarrow e^{y} x = \tan y + C \Rightarrow x = (\tan y + C) e^{- y} Hence, x = (\tan y + C) e^{- y} is the required solution .$

$(xi) We have, x \log x \frac{d y}{d x} + y = 2 \log x Dividing both sides by x \log x, we get \frac{d y}{d x} + \frac{y}{x \log x} = 2 \frac{\log x}{x \log x} \Rightarrow \frac{d y}{d x} + \frac{y}{x \log x} = \frac{2}{x} \Rightarrow \frac{d y}{d x} + (\frac{1}{x \log x}) y = \frac{2}{x} Comparing with \frac{d y}{d x} + P y = Q, we get P = \frac{1}{x \log x} Q = \frac{2}{x} Now, I . F . = e^{\int P d x} = e^{\int \frac{1}{x \log x} d x} = e^{\log |\log x|} = \log x So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow y \log x = 2 \int \frac{1}{x} \times \log x d x + C Putting \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ y \log x = 2 \int t d t + C \Rightarrow y \log x = \frac{2 t^{2}}{2} + C \Rightarrow y \log x = t^{2} + C \Rightarrow y \log x = {(\log x)}^{2} + C (∵ \log x = t) \Rightarrow y = \log x + \frac{C}{\log x}$

$(xii) We have, x \frac{d y}{d x} + 2 y = x^{2} \log x Dividing both sides by x, we get \frac{d y}{d x} + \frac{2 y}{x} = x \log x Comparing with \frac{d y}{d x} + P y = Q, we get P = \frac{2}{x} Q = x \log x Now, I . F . = e^{\int P d x} = e^{\int \frac{2}{x} d x} = e^{2 \log |x|} = x^{2} So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow x^{2} y = \int \underset{II}{x^{3}} \underset{I}{\log x} d x + C \Rightarrow x^{2} y = \log x \int x^{3} d x - \int [\frac{d}{d x} (\log x) \int x^{3} d x] d x + C \Rightarrow x^{2} y = \frac{x^{4} \log x}{4} - \int \frac{x^{3}}{4} d x + C \Rightarrow x^{2} y = \frac{x^{4} \log x}{4} - \frac{x^{4}}{16} + C \Rightarrow y = \frac{x^{2} \log x}{4} - \frac{x^{2}}{16} + \frac{C}{x^{2}} \Rightarrow y = \frac{x^{2}}{16} (4 \log x - 1) + \frac{C}{x^{2}}$

Page No 22.107:

Question 37:

Solve each of the following initial value problems:
(i) $y' + y = e^{x}, y (0) = \frac{1}{2}$

(ii) $x \frac{d y}{d x} - y = \log x, y (1) = 0$

(iii) $\frac{d y}{d x} + 2 y = e^{- 2 x} \sin x, y (0) = 0$

(iv) $x \frac{d y}{d x} - y = (x + 1) e^{- x}, y (1) = 0$

(v) $(1 + y^{2}) d x + (x - e^{- \tan^{- 1} y}) d x = 0, y (0) = 0$

(vi) $\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, y (0) = 1$

(vii) $x \frac{d y}{d x} + y = x \cos x + \sin x, y (\frac{π}{2}) = 1$

(viii) $\frac{d y}{d x} + y \cot x = 4 x cosec x, y (\frac{π}{2}) = 0$

(ix) $\frac{d y}{d x} + 2 y \tan x = \sin x; y = 0 when x = \frac{π}{3}$

(x) $\frac{d y}{d x} - 3 y \cot x = \sin 2 x; y = 2 when x = \frac{π}{2}$
(xi) $\frac{d y}{d x} + y \cot x = 2 \cos x, y (\frac{π}{2}) = 0$
(xii) $d y = \cos x (2 - y \cos e c x) d x$
(xiii) $\tan x (\frac{d y}{d x}) = 2 x \tan x + x^{2} - y; \tan x \neq 0$ given that y = 0 when $x = \frac{π}{2}$ .

Answer:

$(i) We have, y' + y = e^{x} \Rightarrow \frac{d y}{d x} + y = e^{x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 1 and Q = e^{x} ∴ I . F . = e^{\int P d x} = e^{\int 1 d x} = e^{x} Multiplying both sides of (1) by I . F . = e^{x}, we get e^{x} (\frac{d y}{d x} + y) = e^{x} e^{x} \Rightarrow e^{x} \frac{d y}{d x} + e^{x} y = e^{2 x} Integrating both sides with respect to x, we get y e^{x} = \int e^{2 x} d x + C \Rightarrow y e^{x} = \frac{e^{2 x}}{2} + C . . . . . (2) Now, y (0) = \frac{1}{2} ∴ \frac{1}{2} e^{0} = \frac{e^{0}}{2} + C \Rightarrow C = 0 Putting the value of C in (2), we get y e^{x} = \frac{e^{2 x}}{2} \Rightarrow e^{x} = \frac{e^{x}}{2} Hence, y = \frac{e^{x}}{2} is the required solution .$

$(ii) We have, x \frac{d y}{d x} - y = \log x \Rightarrow \frac{d y}{d x} - \frac{y}{x} = \frac{\log x}{x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{1}{x} and Q = \frac{\log x}{x} ∴ I . F . = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x} Multiplying both sides of (1) by I . F . = \frac{1}{x}, we get \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = \frac{1}{x} \times \frac{\log x}{x} \Rightarrow \frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2}} y = \frac{\log x}{x^{2}} Integrating both sides with respect to x, we get y \frac{1}{x} = \int \underset{II}{\frac{1}{x^{2}}} \underset{I}{\times \log x} d x + C \Rightarrow \frac{y}{x} = \log x \int \frac{1}{x^{2}} d x - \int [\frac{d}{d x} (\log x) \int \frac{1}{x^{2}} d x] d x + C \Rightarrow \frac{y}{x} = - \frac{\log x}{x} + \int \frac{1}{x^{2}} d x + C \Rightarrow \frac{y}{x} = - \frac{\log x}{x} - \frac{1}{x} + C \Rightarrow y = - \log x - 1 + C x . . . . . (2) Now, y (1) = 0 ∴ 0 = - 0 - 1 + C (1) \Rightarrow C = 1 Putting the value of C in (2), we get y = - \log x - 1 + x \Rightarrow y = x - 1 - \log x Hence, y = x - 1 - \log x is the required solution .$

$(iii) We have, \frac{d y}{d x} + 2 y = e^{- 2 x} \sin x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 and Q = e^{- 2 x} \sin x ∴ I . F . = e^{\int P d x} = e^{\int 2 d x} = e^{2 x} Multiplying both sides of (1) by I . F . = e^{2 x}, we get e^{2 x} (\frac{d y}{d x} + 2 y) = e^{2 x} e^{- 2 x} \sin x \Rightarrow e^{2 x} (\frac{d y}{d x} + 2 y) = \sin x Integrating both sides with respect to x, we get y e^{2 x} = \int \sin x d x + C \Rightarrow y e^{2 x} = - \cos x + C . . . . . (2) Now, y (0) = 0 ∴ 0 \times e^{0} = - \cos 0 + C \Rightarrow C = 1 Putting the value of C in (2), we get y e^{2 x} = - \cos x + 1 \Rightarrow y e^{2 x} = 1 - \cos x Hence, y e^{2 x} = 1 - \cos x is the required solution .$

$(iv) We have, x \frac{d y}{d x} - y = (x + 1) e^{- x} \Rightarrow \frac{d y}{d x} - \frac{1}{x} y = (\frac{x + 1}{x}) e^{- x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{1}{x} and Q = \frac{x + 1}{x} e^{- x} ∴ I . F . = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x} Multiplying both sides of (1) by I . F . = \frac{1}{x}, we get \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = \frac{1}{x} (\frac{x + 1}{x}) e^{- x} \Rightarrow \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = (\frac{x + 1}{x^{2}}) e^{- x} Integrating both sides with respect to x, we get \frac{1}{x} y = \int (\frac{1}{x} + \frac{1}{x^{2}}) e^{- x} d x + C Putting \frac{1}{x} e^{- x} = t \Rightarrow (- \frac{1}{x} e^{- x} - \frac{1}{x^{2}} e^{- x}) d x = d t \Rightarrow (\frac{1}{x} + \frac{1}{x^{2}}) e^{- x} d x = - d t ∴ \frac{1}{x} y = \int - d t + C \Rightarrow \frac{y}{x} = - t + C \Rightarrow \frac{y}{x} = - \frac{e^{- x}}{x} + C \Rightarrow y = - e^{- x} + C x . . . . . (2) Now, y (1) = 0 ∴ 0 = - e^{- 1} + C \Rightarrow C = e^{- 1} Putting the value of C in (2), we get y = - e^{- x} + x e^{- 1} \Rightarrow y = x e^{- 1} - e^{- x} Hence, y = x e^{- 1} - e^{- x} is the required solution .$

$(v) We have, (1 + y^{2}) d x + (x - e^{- \tan^{- 1} y}) d y = 0 \Rightarrow (x - e^{- \tan^{- 1} y}) \frac{d y}{d x} = - (1 + y^{2}) \Rightarrow (1 + y^{2}) \frac{d x}{d y} = - (x - e^{- \tan^{- 1} y}) \Rightarrow \frac{d x}{d y} + \frac{x}{1 + y^{2}} = \frac{e^{- \tan^{- 1} y}}{1 + y^{2}} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = \frac{1}{1 + y^{2}} and Q = \frac{e^{- \tan^{- 1} y}}{1 + y^{2}} ∴ I . F . = e^{\int P d y} = e^{\int \frac{1}{1 + y^{2}} d y} = e^{\tan^{- 1} y} Multiplying both sides of (1) by I . F . = e^{\tan^{- 1} y}, we get e^{\tan^{- 1} y} (\frac{d x}{d y} + \frac{x}{1 + y^{2}}) = e^{\tan^{- 1} y} \frac{e^{- \tan^{- 1} y}}{1 + y^{2}} \Rightarrow e^{\tan^{- 1} y} (\frac{d x}{d y} + \frac{x}{1 + y^{2}}) = \frac{1}{1 + y^{2}} Integrating both sides with respect to y, we get e^{\tan^{- 1} y} x = \int \frac{1}{1 + y^{2}} d y + C \Rightarrow x e^{\tan^{- 1} y} = \tan^{- 1} y + C . . . . . (2) Now, y (0) = 0 ∴ 0 \times e^{0} = 0 + C \Rightarrow C = 0 Putting the value of C in (2), we get x e^{\tan^{- 1} y} = \tan^{- 1} y + 0 \Rightarrow x e^{\tan^{- 1} y} = \tan^{- 1} y Hence, x e^{\tan^{- 1} y} = \tan^{- 1} y is the required solution .$

$(vi) We have, \frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \tan x and Q = x^{2} \cot x + 2 x ∴ I . F . = e^{\int P d x} = e^{\int \tan x d x} = e^{\log |\sec x|} = \sec x Multiplying both sides of (1) by I . F . = \sec x, we get \sec x (\frac{d y}{d x} + y \tan x) = \sec x (x^{2} \tan x + 2 x) \Rightarrow \sec x (\frac{d y}{d x} + y \tan x) = x^{2} \tan x \sec x + 2 x \sec x Integrating both sides with respect to x, we get y \sec x = \int x^{2} \tan x \sec x d x + 2 \int \underset{II}{x} \underset{I}{\sec x} d x + C \Rightarrow y \sec x = \int x^{2} \tan x \sec x d x + 2 s e c x \int x d x - 2 \int [\frac{d}{d x} (s e c x) \int x d x] d x + C \Rightarrow y \sec x = \int x^{2} \tan x \sec x d x + x^{2} \sec x - \int x^{2} \tan x \sec x d x + C \Rightarrow y \sec x = x^{2} \sec x + C \Rightarrow y = x^{2} + C \cos x . . . . . (2) Now, y (0) = 1 ∴ 1 = 0 + C \cos 0 \Rightarrow C = 1 Putting the value of C in (2), we get y = x^{2} + \cos x Hence, y = x^{2} + \cos x is the required solution .$

$(vii) We have, x \frac{d y}{d x} + y = x \cos x + \sin x \Rightarrow \frac{d y}{d x} + \frac{1}{x} y = \cos x + \frac{\sin x}{x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \tan x and Q = x^{2} \cot x + 2 x ∴ I . F . = e^{\int P d x} = e^{\int \frac{1}{x} d x} = e^{\log x} = x Multiplying both sides of (1) by I . F . = x, we get x (\frac{d y}{d x} + \frac{1}{x} y) = x (\cos x + \frac{\sin x}{x}) \Rightarrow x (\frac{d y}{d x} + \frac{1}{x} y) = x \cos x + \sin x Integrating both sides with respect to x, we get x y = \int x \cos x d x + \int \sin x d x + C \Rightarrow x y = [x \sin x - \int 1 (\sin x) d x] - \cos x + C \Rightarrow x y = x \sin x + \cos x - \cos x + C \Rightarrow x y = x \sin x + C . . . . . (2) Now, y (\frac{π}{2}) = 1 ∴ 1 \times \frac{π}{2} = \frac{π}{2} \sin \frac{π}{2} + C \Rightarrow C = 0 Putting the value of C in (2), we get x y = x \sin x \Rightarrow y = \sin x Hence, y = \sin x is the required solution .$

$(viii) We have, \frac{d y}{d x} + y \cot x = 4 x cosec x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cot x and Q = 4 x cosec x ∴ I . F . = e^{\int P d x} = e^{\int \cot x d x} = e^{\log |\sin x|} = \sin x Multiplying both sides of (1) by I . F . = \sin x, we get \sin x (\frac{d y}{d x} + y \cot x) = \sin x (4 x cosec x) \Rightarrow \sin x (\frac{d y}{d x} + y \cot x) = 4 x Integrating both sides with respect to x, we get y \sin x = 4 \int x d x + C \Rightarrow y \sin x = 2 x^{2} + C . . . . . (2) Now, y (\frac{π}{2}) = 0 ∴ 0 \times \sin (\frac{π}{2}) = 2 {(\frac{π}{2})}^{2} + C \Rightarrow C = - \frac{π^{2}}{2} Putting the value of C in (2), we get y \sin x = 2 x^{2} - \frac{π^{2}}{2} Hence, y \sin x = 2 x^{2} - \frac{π^{2}}{2} is the required solution .$

$(ix) We have, \frac{d y}{d x} + 2 y \tan x = \sin x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 \tan x and Q = \sin x ∴ I . F . = e^{\int P d x} = e^{2 \int \tan x d x} = e^{2 \log |\sec x|} = \sec^{2} x Multiplying both sides of (1) by I . F . = \sec^{2} x, we get \sec^{2} x (\frac{d y}{d x} + 2 y \tan x) = \sec^{2} x \times \sin x \Rightarrow \sec^{2} x (\frac{d y}{d x} + 2 y \tan x) = \tan x \sec x Integrating both sides with respect to x, we get y \sec^{2} x = \int \tan x \sec x d x + C \Rightarrow y \sec^{2} x = \sec x + C . . . . . (2) Now, y (\frac{π}{3}) = 0 ∴ 0 {(\sec \frac{π}{3})}^{2} = \sec \frac{π}{3} + C \Rightarrow C = - 2 Putting the value of C in (2), we get y \sec^{2} x = \sec x - 2 \Rightarrow y = \cos x - 2 \cos^{2} x Hence, y = \cos x - 2 \cos^{2} x is the required solution .$

$(x) We have, \frac{d y}{d x} - 3 y \cot x = \sin 2 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - 3 \cot x and Q = \sin 2 x ∴ I . F . = e^{\int P d x} = e^{- 3 \int \cot x d x} = e^{- 3 \log |\sin x|} = {cosec}^{3} x Multiplying both sides of (1) by I . F . = {cosec}^{3} x, we get {cosec}^{3} x (\frac{d y}{d x} - 3 y \cot x) = \sin 2 x ({cosec}^{3} x) \Rightarrow {cosec}^{3} x (\frac{d y}{d x} - 3 y \cot x) = 2 \cot x cosec x Integrating both sides with respect to x, we get y {cosec}^{3} x = 2 \int \cot x cosec x d x + C \Rightarrow y {cosec}^{3} x = - 2 cosec x + C \Rightarrow y = - 2 \sin^{2} x + C \sin^{3} x . . . . . (2) Now, y (\frac{π}{2}) = 2 ∴ 2 = - 2 \sin^{2} \frac{π}{2} + C \sin^{3} \frac{π}{2} \Rightarrow C = 4 Putting the value of C in (2), we get y = - 2 \sin^{2} x + 4 \sin^{3} x \Rightarrow y = 4 \sin^{3} x - 2 \sin^{2} x Hence, y = 4 \sin^{3} x - 2 \sin^{2} x is the required solution .$

$(xi) \frac{d y}{d x} + y \cot x = 2 \cos x, y (\frac{π}{2}) = 0 \frac{d y}{d x} + y \cot x = 2 \cos x . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cot x and Q = 2 \cos x ∴ I . F . = e^{\int P d x} = e^{\int \cot x d x} = e^{logsin x} = \sin x Multiplying both sides of (1) by I . F . = \sin x, we get \sin x (\frac{d y}{d x} + y \cot x) = 2 \sin x \cos x \Rightarrow \sin x \frac{d y}{d x} + y \cos x = \sin 2 x Integrating both sides with respect to x, we get y \sin x = \int \sin 2 x d x + C \Rightarrow y \sin x = - \frac{\cos 2 x}{2} + C . . . . . (2) Now, y (\frac{π}{2}) = 0 ∴ 0 \times \sin (\frac{π}{2}) = - \frac{cosπ}{2} + C \Rightarrow C = - \frac{1}{2} Putting the value of C in (2), we get y \sin x = - \frac{\cos 2 x}{2} - \frac{1}{2} \Rightarrow 2 y \sin x = - (1 + \cos 2 x) \Rightarrow 2 y \sin x = - 2 \cos^{2} x \Rightarrow y = - \cot x \cos x Hence, y = - \cot x \cos x is the required solution .$

$(x i i) d y = \cos x (2 - y \cos e c x) d x \Rightarrow \frac{d y}{d x} = 2 \cos x - y c o t x \Rightarrow \frac{d y}{d x} + y \cot x = 2 \cos x . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cot x and Q = 2 \cos x ∴ I . F . = e^{\int P d x} = e^{\int \cot x d x} = e^{logsin x} = \sin x Multiplying both sides of (1) by I . F . = \sin x, we get \sin x (\frac{d y}{d x} + y \cot x) = 2 \sin x \cos x \Rightarrow \sin x \frac{d y}{d x} + y \cos x = \sin 2 x Integrating both sides with respect to x, we get y \sin x = \int \sin 2 x d x + C \Rightarrow y \sin x = - \frac{\cos 2 x}{2} + C Hence, y \sin x = - \frac{\cos 2 x}{2} + C is the required solution .$

(xiii)
$\tan x (\frac{d y}{d x}) = 2 x \tan x + x^{2} - y \Rightarrow \frac{d y}{d x} + \frac{1}{\tan x} y = \frac{2 x \tan x + x^{2}}{\tan x} \Rightarrow \frac{d y}{d x} + (\cot x) y = 2 x + x^{2} \cot x$
This is a linear differential equation of the form $\frac{d y}{d x} + P y = Q$ .

Integrating factor, I.F. = $e^{\int P d x} = e^{\int \cot x d x} = e^{\log \sin x} = \sin x$

The solution of the given differential equation is given by

$y \times (I . F .) = \int Q \times (I . F .) d x + C \Rightarrow y \times \sin x = \int (2 x + x^{2} \cot x) \sin x d x + C \Rightarrow y \sin x = \int 2 x \sin x d x + \int x^{2} \cos x d x + C \Rightarrow y \sin x = \int 2 x \sin x d x + [x^{2} \int \cos x d x - \int (\frac{d}{d x} x^{2} \times \int \cos x d x) d x] + C$
$\Rightarrow y \sin x = \int 2 x \sin x d x + x^{2} \sin x - \int 2 x \sin x d x + C \Rightarrow y \sin x = x^{2} \sin x + C \Rightarrow y = x^{2} + cosec x \times C . . . . . (1)$
It is given that, y = 0 when $x = \frac{π}{2}$ .

$∴ 0 = {(\frac{π}{2})}^{2} + cosec \frac{π}{2} \times C \Rightarrow C = - \frac{π^{2}}{4}$
Putting $C = - \frac{π^{2}}{4}$ in (1), we get

$y = x^{2} - \frac{π^{2}}{4} cosec x$

Hence, $y = x^{2} - \frac{π^{2}}{4} cosec x$ is the required solution.

Page No 22.107:

Question 38:

Find the general solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2}$ .

Answer:

$We have, x \frac{d y}{d x} + 2 y = x^{2} \Rightarrow \frac{d y}{d x} + \frac{2}{x} y = x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \frac{2}{x} and Q = x . ∴ I . F . = e^{\int P d x} = e^{\int \frac{2}{x} d x} = e^{2 \log x} = x^{2} Multiplying both sides of (1) by I . F . = x^{2}, we get x^{2} (\frac{d y}{d x} + \frac{2}{x} y) = x^{2} x \Rightarrow x^{2} \frac{d y}{d x} + 2 x y = x^{3} Integrating both sides with respect to x, we get x^{2} y = \int x^{3} d x + C \Rightarrow x^{2} y = \frac{x^{4}}{4} + C \Rightarrow y = \frac{x^{2}}{4} + C x^{- 2} Hence, y = \frac{x^{2}}{4} + C x^{- 2} is the required solution .$

Page No 22.107:

Question 39:

Find the general solution of the differential equation $\frac{d y}{d x} - y = \cos x$ .

Answer:

$We have, \frac{d y}{d x} - y = \cos x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - 1 and Q = \cos x ∴ I . F . = e^{\int P d x} = e^{- \int d x} = e^{- x} Multiplying both sides of (1) by I . F . = e^{- x}, we get e^{- x} (\frac{d y}{d x} - y) = e^{- x} \cos x \Rightarrow e^{- x} \frac{d y}{d x} - e^{- x} y = e^{- x} \cos x Integrating both sides with respect to x, we get y e^{- x} = \int e^{- x} \cos x d x + C \Rightarrow y e^{- x} = I + C . . . . . (2) Here, I = \int e^{- x} \cos x d x . . . . . (3) \Rightarrow I = e^{- x} \sin x - \int (- e^{- x} \sin x) d x \Rightarrow I = e^{- x} \sin x + \int e^{- x} \sin x d x \Rightarrow I = e^{- x} \sin x - e^{- x} \cos x - \int [(- e^{- x}) \times (- \cos x)] d x \Rightarrow I = e^{- x} \sin x - e^{- x} \cos x - \int e^{- x} \cos x d x \Rightarrow I = e^{- x} \sin x - e^{- x} \cos x - I [From (3)] \Rightarrow 2 I = e^{- x} (\sin x - \cos x) \Rightarrow I = \frac{e^{- x}}{2} (\sin x - \cos x) . . . . . (4) From (2) and (4) we get \Rightarrow y e^{- x} = \frac{e^{- x}}{2} (\sin x - \cos x) + C \Rightarrow y = \frac{1}{2} (\sin x - \cos x) + C e^{x} Hence, y = \frac{1}{2} (\sin x - \cos x) + C e^{x} is the required solution .$

Page No 22.107:

Question 40:

Solve the differential equation $(y + 3 x^{2}) \frac{d x}{d y} = x$

Answer:

$We have, (y + 3 x^{2}) \frac{d x}{d y} = x \Rightarrow \frac{d y}{d x} = \frac{y + 3 x^{2}}{x} \Rightarrow \frac{d y}{d x} - \frac{1}{x} y = 3 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{1}{x} and Q = 3 x ∴ I . F . = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x} Multiplying both sides of (1) by I . F . = \frac{1}{x}, we get \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = \frac{1}{x} 3 x \Rightarrow \frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2}} y = 3 Integrating both sides with respect to x, we get \frac{1}{x} y = 3 \int d x + C \Rightarrow \frac{y}{x} = 3 x + C Hence, \frac{y}{x} = 3 x + C is the required solution .$

Page No 22.108:

Question 41:

Find the particular solution of the differential equation $\frac{d x}{d y} + x \cot y = 2 y + y^{2}$ cot y, y ≠ 0 given that x = 0 when $y = \frac{π}{2}$ .

Answer:

$We have, \frac{d x}{d y} + x \cot y = 2 y + y^{2} \cot y . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = \cot y and Q = 2 y + y^{2} \cot y ∴ I . F . = e^{\int P d y} = e^{\int \cot y d y} = e^{\log |\sin y|} = \sin y Multiplying both sides of (1) by I . F . = \sin y, we get \sin y (\frac{d x}{d y} + x \cot y) = \sin y (y^{2} \cot y + 2 y) \Rightarrow \sin y \frac{d x}{d y} + x \cos y = y^{2} \cos y + 2 y \sin y Integrating both sides with respect to y, we get x \sin y = \int \underset{I}{y^{2}} \underset{II}{\cos y} d y + \int 2 y \sin y d y + C \Rightarrow x \sin y = y^{2} \int \cos y d y - \int [\frac{d}{d y} (y^{2}) \int \cos y d y] d y + \int 2 y \sin y d y + C \Rightarrow x \sin y = y^{2} \sin y - \int 2 y \sin y d y + \int 2 y \sin y d y + C \Rightarrow x \sin y = y^{2} \sin y + C Now, y = \frac{π}{2} at x = 0 ∴ 0 \times \sin \frac{π}{2} = {\frac{π}{4}}^{2} \sin \frac{π}{2} + C \Rightarrow C = - {\frac{π}{4}}^{2} Putting the value of C, we get x \sin y = y^{2} \sin y - {\frac{π}{4}}^{2} Hence, x \sin y = y^{2} \sin y - {\frac{π}{4}}^{2} is the required solution .$

Page No 22.108:

Question 42:

Solve the following differential equation:

$(\cot^{- 1} y + x) d y = (1 + y^{2}) d x$

Answer:

The given differential equation is $(\cot^{- 1} y + x) d y = (1 + y^{2}) d x$ .
This differential equation can be written as
$\frac{d x}{d y} = \frac{\cot^{- 1} y + x}{1 + y^{2}} \Rightarrow \frac{d x}{d y} + (- \frac{1}{1 + y^{2}}) x = \frac{\cot^{- 1} y}{1 + y^{2}}$
This is a linear differential equation with $P = - \frac{1}{1 + y^{2}} and Q = \frac{\cot^{- 1} y}{1 + y^{2}}$ .
I.F. = $e^{- \int \frac{1}{1 + y^{2}} d y} = e^{\cot^{- 1} y}$
Multiply the differential equation by integration factor (I.F.), we get
$\frac{d x}{d y} e^{\cot^{- 1} y} - \frac{x}{(1 + y^{2})} e^{\cot^{- 1} y} = \frac{\cot^{- 1} y}{(1 + y^{2})} e^{\cot^{- 1} y} \Rightarrow \frac{d}{d y} (x e^{\cot^{- 1} y}) = \frac{\cot^{- 1} y}{(1 + y^{2})} e^{\cot^{- 1} y}$
Integrating both sides with respect y, we get
$x e^{\cot^{- 1} y} = \int \frac{\cot^{- 1} y}{(1 + y^{2})} e^{\cot^{- 1} y} d y + C$
Putting $t = \cot^{- 1} y$ and $d t = - \frac{1}{1 + y^{2}} d y$ , we get
$x e^{\cot^{- 1} y} = - \int t e^{t} d t + C \Rightarrow x e^{\cot^{- 1} y} = - e^{t} (t - 1) + C \Rightarrow x e^{\cot^{- 1} y} = e^{\cot^{- 1} y} (1 - \cot^{- 1} y) + C$

Page No 22.134:

Question 1:

The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.

Answer:

Let r be the radius and S be the surface area of the balloon at any time t. Then,
$S = 4 π r^{2} \Rightarrow \frac{d S}{d t} = 8 π r \frac{d r}{d t} . . . . . (1) Given : \frac{d S}{d t} α t \Rightarrow \frac{d S}{d t} = k t, where k is any constant Putting \frac{d S}{d t} = k t in (1), we get \Rightarrow k t = 8 π r \frac{d r}{d t} k t d t = 8 π r d r Integrating both sides, we get \int k t d t = \int 8 π r d r \Rightarrow \frac{k t^{2}}{2} = 8 π \times \frac{r^{2}}{2} + C . . . . . (2) At t = 0 s, r = 1 unit and at t = 3 s, r = 2 units (Given) ∴ 0 = 8 π \times \frac{1}{2} + C \Rightarrow C = - 4 π And \frac{9}{2} k = 8 π \times 2 + C \Rightarrow \frac{9}{2} k = 12 π \Rightarrow k = \frac{8}{3} π Substituting the values of C and k in (2), we get \frac{8 t^{2}}{6} π = 8 π \times \frac{r^{2}}{2} - 4 π \Rightarrow \frac{4 t^{2}}{3} = 4 r^{2} - 4 \Rightarrow \frac{t^{2}}{3} = r^{2} - 1 \Rightarrow r^{2} = 1 + \frac{t^{2}}{3} \Rightarrow r = \sqrt{1 + \frac{1}{3} t^{2}}$

Page No 22.134:

Question 2:

A population grows at the rate of 5% per year. How long does it take for the population to double?

Answer:

Let P₀be the initial population and P be the population at any time t. Then,
$\frac{d P}{d t} = \frac{5 P}{100} \Rightarrow \frac{d P}{d t} = 0.05 P$

$\Rightarrow \frac{d P}{P} = 0.05 d t Integrating both sides with respect to t, we get \int \frac{d P}{P} = \int 0.05 d t \log P = 0.05 t + C Now, P = P_{0} at t = 0 ∴ \log P_{0} = 0 + C \Rightarrow C = \log P_{0} Putting the value of C, we get \log P = 0.05 t + \log P_{0} \Rightarrow \log \frac{P}{P_{0}} = 0.05 t To find the time when the population will double, we have P = 2 P_{0} ∴ \log \frac{2 P_{0}}{P_{0}} = 0.05 t \Rightarrow \log 2 = 0.05 t \Rightarrow t = \frac{\log 2}{0.05} = 20 \log 2 years$

Page No 22.134:

Question 3:

The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?

Answer:

Let the original population be N and the population at any time t be P.
Given: $\frac{d P}{d t} α P$
$\Rightarrow \frac{d P}{d t} = a P \Rightarrow \frac{d P}{P} = a d t \Rightarrow \log |P| = a t + C . . . . . (1) Now, P = N at t = 0 Putting P = N and t = 0 in (1), we get \log |N| = C Putting C = \log |N| in (1), we get \log |P| = a t + \log |N| \Rightarrow \log |\frac{P}{N}| = a t . . . . . (2) According to the question, \log |\frac{2 N}{N}| = 25 a \Rightarrow a = \frac{1}{25} \log |2| = \frac{1}{25} \times 0.6931 = 0.0277 Putting a = 0.0277 in (2), we get \log |\frac{P}{N}| = 0.0277 t . . . (3) For P = 500000 and N = 100000 : \log |\frac{500000}{100000}| = 0.0277 t \Rightarrow t = \frac{\log 5}{0.0277} = \frac{1.609}{0.0277} = 58.08 years = Approximately 58 years$

Page No 22.134:

Question 4:

In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?

Answer:

$Let the bacteria count at any time t be N . Given : \frac{d N}{d t} α N \Rightarrow \frac{d N}{d t} = λ N \Rightarrow \frac{1}{N} d N = λ d t Integrating both sides, we get \int \frac{1}{N} d N = \int λ d t \Rightarrow \log N = λ t + \log C . . . . . (1) Initially when t = 0, then N = 100000 (Given) ∴ \log 100000 = 0 + \log C \Rightarrow \log C = \log 100000 After 2 hours number increased by 10 % Therefore, increased number = 100000 (1 + 10 %) = 110000 Given : t = 2, N = 110000 Putting t = 2, N = 110000 in (1), we get \log 110000 = 2 λ + \log 100000 \Rightarrow \frac{1}{2} \log \frac{11}{10} = λ Substituting the values of \log C and λ in (1), we get \log N = \frac{t}{2} \log (\frac{11}{10}) + \log 100000 . . . . . (2) Now, Let t = T when N = 200000 Substituting these values in (2), we get \log 200000 = \frac{T}{2} \log (\frac{11}{10}) + \log 100000 \Rightarrow \log 2 = \frac{T}{2} \log \frac{11}{10} \Rightarrow T = 2 \frac{\log 2}{\log \frac{11}{10}} ∴ The count will reach 200000 in 2 \frac{\log 2}{\log \frac{11}{10}} hours .$

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Question 5:

If the interest is compounded continuously at 6% per annum, how much worth Rs 1000 will be after 10 years? How long will it take to double Rs 1000?

Answer:

Let P₀be the initial amount and P be the amount at any time t. Then,
$\frac{d P}{d t} = \frac{6 P}{100} \Rightarrow \frac{d P}{d t} = 0.06 P$
$\Rightarrow \frac{d P}{P} = 0.06 d t Integrating both sides with respect to t, we get \log P = 0.06 t + C Now, P = P_{0} at t = 0 ∴ \log P_{0} = 0 + C \Rightarrow C = \log P_{0} Putting the value of C, we get \log P = 0.06 t + \log P_{0} \Rightarrow \log \frac{P}{P_{0}} = 0.06 t \Rightarrow e^{0.06 t} = \frac{P}{P_{0}} To find the amount after 10 years, we get \Rightarrow e^{0.06 \times 10} = \frac{P}{P_{0}} \Rightarrow e^{0.6} = \frac{P}{P_{0}} \Rightarrow 1.822 = \frac{P}{P_{0}} \Rightarrow P = 1.822 P_{0} \Rightarrow P = 1.822 \times 1000 = Rs 1822 To find the time after which the amount will double, we have P = 2 P_{0} ∴ \log \frac{2 P_{0}}{P_{0}} = 0.06 t \Rightarrow \log 2 = 0.06 t \Rightarrow t = \frac{0.6931}{0.06} = 11.55 years$

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Question 6:

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.

Answer:

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: $\frac{d P}{d t} α P$
$\Rightarrow \frac{d P}{d t} = a P \Rightarrow \frac{d P}{P} = a d t \Rightarrow \log |P| = a t + C . . . . . (1) Now, P = N at t = 0 Putting P = N and t = 0 in (1), we get \log |N| = C Putting C = \log |N| in (1), we get \log |P| = a t + \log |N| \Rightarrow \log |\frac{P}{N}| = a t . . . . . (2) According to the question, \log |\frac{3 N}{N}| = 5 a \Rightarrow a = \frac{1}{5} \log |3| = \frac{1}{5} \times 1.0986 = 0.21972 Putting a = 0.21972 in (2), we get \log |\frac{P}{N}| = 0.21972 t . . . . . (3) \Rightarrow e^{0.21972 t} = \frac{P}{N} . . . . . (4) Putting t = 10 in (4) to find the bacteria after 10 hours, we get e^{0.21972 \times 10} = \frac{P}{N} \Rightarrow e^{2.1972} = \frac{P}{N} \Rightarrow \frac{P}{N} = 9 \Rightarrow P = 9 N To find the time taken when the number of bacteria becomes 10 times of the number of initial population, we have P = 10 N ∴ \log |\frac{10 N}{N}| = \frac{1}{5} t \log 3 \Rightarrow t = \frac{5 \log 10}{\log 3}$

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Question 7:

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?

Answer:

Let the population at any time t be P.

Given: $\frac{d P}{d t} α P$

$\Rightarrow \frac{d P}{d t} = β P \Rightarrow \frac{d P}{P} = β d t \Rightarrow \log |P| = β t + \log C . . . (1) Now, At t = 1990, P = 200000 and at t = 2000, P = 250000 ∴ \log 200000 = 1990 β + \log C . . . (2) \log 250000 = 2000 β + \log C . . . (3) Subtracting (3) from (2), we get \log 200000 - \log 250000 = 10 β \Rightarrow β = \frac{1}{10} \log (\frac{5}{4}) Putting β = \frac{1}{10} \log (\frac{5}{4}) in (2), we get \log 200000 = 1990 \times \frac{1}{10} \log (\frac{5}{4}) + \log C \Rightarrow \log 200000 = 199 \log (\frac{5}{4}) + \log C \Rightarrow \log C = \log 200000 - 199 \log (\frac{5}{4}) Putting β = \frac{1}{10} \log (\frac{5}{4}), \log C = \log 200000 - 199 \log (\frac{5}{4}) and t = 2010 in (1), we get \log |P| = \frac{1}{10} \times 2010 \log (\frac{5}{4}) + \log 200000 - 199 \log (\frac{5}{4}) \Rightarrow \log |P| = 201 \log (\frac{5}{4}) + \log 200000 - 199 \log (\frac{5}{4}) \Rightarrow \log |P| = \log {(\frac{5}{4})}^{201} - \log {(\frac{5}{4})}^{199} + \log 200000 \Rightarrow \log |P| = \log \{{(\frac{5}{4})}^{201} {(\frac{4}{5})}^{199}\} + \log 200000 \Rightarrow \log |P| = \log \{{(\frac{5}{4})}^{2}\} + \log 200000 \Rightarrow \log |P| = \log (\frac{25}{16} \times 200000) \Rightarrow \log |P| = \log 312500 \Rightarrow P = 312500$

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Question 8:

If the marginal cost of manufacturing a certain item is given by C' (x) = $\frac{d C}{d x}$ = 2 + 0.15 x. Find the total cost function C (x), given that C (0) = 100.

Answer:

$We have, \frac{d C}{d x} = 2 + 0.15 x \Rightarrow d C = (2 + 0.15 x) d x Integrating both sides with respect to x, we get C = 2 x + \frac{0.15}{2} x^{2} + K . . . . . (1) At C (0) = 100, we have 100 = 2 (0) + \frac{0.15}{2} {(0)}^{2} + K \Rightarrow K = 100 Putting the value of T in (1), we get C = 2 x + \frac{0.15}{2} x^{2} + 100 \Rightarrow C = 0.075 x^{2} + 2 x + 100$

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Question 9:

A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.

Answer:

Let P₀be the initial amount and P be the amount at any time t.
We have,
$\frac{d P}{d t} = \frac{8 P}{100} \Rightarrow \frac{d P}{d t} = \frac{2 P}{25}$

$\Rightarrow \frac{d P}{P} = \frac{2}{25} d t Integrating both sides with respect to t, we get \log P = \frac{2}{25} t + C . . . . . (1) Now, P = P_{0} at t = 0 ∴ \log P_{0} = 0 + C \Rightarrow C = \log P_{0} Putting the value of C in (1), we get \log P = \frac{2}{25} t + \log P_{0} \Rightarrow \log \frac{P}{P_{0}} = \frac{2}{25} t \Rightarrow e^{\frac{2}{25} t} = \frac{P}{P_{0}} To find the amount after 1 year, we have e^{\frac{2}{25}} = \frac{P}{P_{0}} \Rightarrow e^{0.08} = \frac{P}{P_{0}} \Rightarrow 1.0833 = \frac{P}{P_{0}} \Rightarrow P = 1.0833 P_{0} Percentage increase = (\frac{P - P_{0}}{P_{0}}) \times 100 % = (\frac{1.0833 P_{0} - P_{0}}{P_{0}}) \times 100 % = 0.0833 \times 100 % = 8.33 %$

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Question 10:

In a simple circuit of resistance R, self inductance L and voltage E, the current i at any time t is given by L $\frac{d i}{d t}$ + R i = E. If E is constant and initially no current passes through the circuit, prove that $i = \frac{E}{R} \{1 - e^{- (R / L) t}\}$ .

Answer:

$We have, L \frac{d i}{d t} + R i = E \Rightarrow \frac{d i}{d t} + \frac{R}{L} i = \frac{E}{L} . . . . . (1) ∴ I . F . = e^{\int \frac{R}{L} d t} = e^{\frac{R}{L} t} Multiplying both sides of (1) by I . F . = e^{\frac{R}{L} t}, we get e^{\frac{R}{L} t} (\frac{d i}{d t} + \frac{R}{L} i) = e^{\frac{R}{L} t} \times \frac{E}{L} \Rightarrow e^{\frac{R}{L} t} \frac{d i}{d t} + e^{\frac{R}{L} t} \frac{R}{L} i = e^{\frac{R}{L} t} \times \frac{E}{L} Integrating both sides with respect to t, we get e^{\frac{R}{L} t} i = \frac{E}{L} \int e^{\frac{R}{L} t} d t + C \Rightarrow e^{\frac{R}{L} t} i = \frac{E}{L} \times \frac{L}{R} e^{\frac{R}{L} t} + C \Rightarrow e^{\frac{R}{L} t} i = \frac{E}{R} e^{\frac{R}{L} t} + C . . . . . (2) Now, i = 0 at t = 0 ∴ e^{0} \times 0 = \frac{E}{R} e^{0} + C \Rightarrow C = - \frac{E}{R} Putting the value of C in (2), we get e^{\frac{R}{L} t} i = \frac{E}{R} e^{\frac{R}{L} t} - \frac{E}{R} \Rightarrow i = \frac{E}{R} - \frac{E}{R} e^{- \frac{R}{L} t} \Rightarrow i = \frac{E}{R} (1 - e^{- \frac{R}{L} t})$

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Question 11:

The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.

Answer:

Let the initial amount of radium be N and the amount of radium present at any time t be P.
Given: $\frac{d P}{d t} α P$
$\Rightarrow \frac{d P}{d t} = - a P, where a > 0 \Rightarrow \frac{d P}{P} = - a d t Integrating both sides, we get \Rightarrow \log |P| = - a t + C . . . . . (1) Now, P = N at t = 0 Putting P = N and t = 0 in (1), we get \log |N| = C Putting C = \log |N| in (1), we get \log |P| = - a t + \log |N| \Rightarrow \log |\frac{N}{P}| = a t According to the question, \log |\frac{N}{\frac{N}{2}}| = a t \Rightarrow \log |2| = a t \Rightarrow t = \frac{1}{a} \log |2| Here, a is the constant of proportionality .$

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Question 12:

Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?

Answer:

Let the original amount of the radium be N and the amount of radium at any time t be P.
Given: $\frac{d P}{d t} α P$
$\Rightarrow \frac{d P}{d t} = - a P \Rightarrow \frac{d P}{P} = - a d t Integrating both sides, we get \Rightarrow \log |P| = - a t + C . . . . . (1) Now, P = N at t = 0 Putting P = N and t = 0 in (1), we get \log |N| = C Putting C = \log |N| in (1), we get \log |P| = - a t + \log |N| \Rightarrow \log |\frac{P}{N}| = - a t . . . . . (2) According to the question, P = \frac{1}{2} N a t t = 1590 \log |\frac{N}{2 N}| = - 1590 a \Rightarrow - \log 2 = - 1590 a \Rightarrow a = \frac{1}{1590} \log 2 Putting a = \frac{1}{1590} \log 2 in (2), we get \log |\frac{P}{N}| = - (\frac{1}{1590} \log 2) t \frac{P}{N} = e^{- \frac{\log 2}{1590} t} . . . . . (3) Putting t = 1 in (4) to find the bacteria after 1 year, we get \frac{P}{N} = 0.9996 \Rightarrow P = 0.9996 N Percentage of amount disapeared in 1 year = (\frac{N - P}{N}) \times 100 % = \frac{N - 0.9996 N}{N} \times 100 % = 0.04 %$

Page No 22.135:

Question 13:

The slope of the tangent at a point P (x, y) on a curve is $\frac{- x}{y}$ . If the curve passes through the point (3, −4), find the equation of the curve.

Answer:

$According to the question, \frac{d y}{d x} = \frac{- x}{y} \Rightarrow y d y = - x d x Integrating both sides with respect to x, we get \int y d y = - \int x d x \Rightarrow \frac{y^{2}}{2} = - \frac{x^{2}}{2} + C Since the curve passes through (3, - 4), it satisfies the above equation . ∴ \frac{{(- 4)}^{2}}{2} = - \frac{3^{2}}{2} + C \Rightarrow 8 = - \frac{9}{2} + C \Rightarrow C = \frac{25}{2} Putting the value of C, we get \frac{y^{2}}{2} = - \frac{x^{2}}{2} + \frac{25}{2} \Rightarrow x^{2} + y^{2} = 25$

Page No 22.135:

Question 14:

Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation $y - x \frac{d y}{d x} = y^{2} + \frac{d y}{d x}$ .

Answer:

We have,
$y - x \frac{d y}{d x} = y^{2} + \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x} (x + 1) = y (1 - y) \Rightarrow \frac{d y}{y (1 - y)} = \frac{d x}{(x + 1)}$

$Integrating both sides, we get \int \frac{d y}{y (1 - y)} = \int \frac{d x}{x + 1} \Rightarrow \int (\frac{1}{y} + \frac{1}{1 - y}) d y = \int \frac{d x}{x + 1} \Rightarrow \log |y| - \log |1 - y| = \log |x + 1| + C . . . . . (1) Since the curve passes throught the point (2, 2), it satisfies the equation of the curve . \Rightarrow \log |2| - \log |1 - 2| = \log |2 + 1| + C \Rightarrow C = \log |\frac{2}{3}| Putting the value of C in (1), we get \log |y| - \log |1 - y| = \log |x + 1| + \log |\frac{2}{3}| \Rightarrow \log |\frac{y}{(1 - y)}| = \log |\frac{2 (x + 1)}{3}| \Rightarrow |\frac{y}{(1 - y)}| = |\frac{2 (x + 1)}{3}| \Rightarrow \frac{y}{(1 - y)} = \pm \frac{2 (x + 1)}{3} \Rightarrow \frac{y}{(1 - y)} = \frac{2 (x + 1)}{3} or \frac{y}{(1 - y)} = - \frac{2 (x + 1)}{3} Here, given point (2, 2) does not satisfy \frac{y}{(1 - y)} = \frac{2 (x + 1)}{3} But it satisfy \frac{y}{(1 - y)} = - \frac{2 (x + 1)}{3} ∴ \frac{y}{(1 - y)} = - \frac{2 (x + 1)}{3} \Rightarrow \frac{y}{(y - 1)} = \frac{2 (x + 1)}{3} \Rightarrow 3 y = 2 (x + 1) (y - 1) \Rightarrow 3 y = 2 x y - 2 x + 2 y - 2 \Rightarrow 2 x y - 2 x - y - 2 = 0$

Page No 22.135:

Question 15:

Find the equation of the curve passing through the point $(1, \frac{π}{4})$ and tangent at any point of which makes an angle tan⁻¹ $(\frac{y}{x} - \cos^{2} \frac{y}{x})$ with x-axis.

Answer:

The slope of the curve is given as $\frac{d y}{d x} = \tan θ$ .
Here,
$θ = \tan^{- 1} (\frac{y}{x} - \cos^{2} \frac{y}{x})$

$∴ \frac{d y}{d x} = \tan \{\tan^{- 1} (\frac{y}{x} - \cos^{2} \frac{y}{x})\} \Rightarrow \frac{d y}{d x} = \frac{y}{x} - \cos^{2} \frac{y}{x}$

$Let y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} ∴ v + x \frac{d v}{d x} = v - \cos^{2} v \Rightarrow x \frac{d v}{d x} = - \cos^{2} v \Rightarrow \sec^{2} v d v = - \frac{1}{x} d x Integrating both sides with respect to x, we get \int \sec^{2} v d v = - \int \frac{1}{x} d x \Rightarrow \tan v = - \log |x| + C \Rightarrow \tan \frac{y}{x} = - \log |x| + C Since the curve passes through (1, \frac{π}{4}), it satisfies the above equation . ∴ \tan \frac{π}{4} = - \log |1| + C \Rightarrow C = 1 Putting the value of C, we get \tan \frac{y}{x} = - \log |x| + 1 \Rightarrow \tan \frac{y}{x} = - \log |x| + \log e \Rightarrow \tan \frac{y}{x} = \log |\frac{e}{x}|$

Page No 22.135:

Question 16:

Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.

Answer:

Let the given curve be y = f(x). Suppose P(x,y) be a point on the curve. Equation of the tangent to the curve at P is
$Y - y = \frac{d y}{d x} (X - x)$ , where (X, Y) is the arbitrary point on the tangent.
Putting Y=0 we get,
$0 - y = \frac{d y}{d x} (X - x) Therefore, X - x = - y \frac{d x}{d y} \Rightarrow X = x - y \frac{d x}{d y} Therefore, cut off by the tangent on the x - axis = x - y \frac{d x}{d y} G i v e n, x - y \frac{d x}{d y} = 4 y T h e r e f o r e, - y \frac{d x}{d y} = 4 y - x \Rightarrow \frac{d x}{d y} = \frac{x - 4 y}{y} \Rightarrow \frac{d y}{d x} = \frac{y}{x - 4 y} . . . . . . . . (1) this is a homogeneous differential equation .$
$Putting y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x} i n (1) we get v + x \frac{d v}{d x} = \frac{v x}{x - 4 v x} T h e r e f o r e, v + x \frac{d v}{d x} = \frac{v}{1 - 4 v} \Rightarrow \frac{x d v}{d x} = \frac{v}{1 - 4 v} - v = \frac{4 v^{2}}{1 - 4 v} \Rightarrow \frac{1 - 4 v}{v^{2}} d v = 4 \frac{d x}{x}$
Integrating on both sides we get,
$\int \frac{1 - 4 v}{v^{2}} d v = 4 \int \frac{d x}{x} T h e r e f o r e, \int \frac{d v}{v^{2}} - 4 \int \frac{d v}{v} = 4 \int \frac{d x}{x} \Rightarrow - \frac{1}{v} - 4 \log v = 4 \log x + \log c \Rightarrow - \frac{1}{v} = 4 l o g x + l o g c + 4 l o g v \Rightarrow 4 \log (x v) + \log c = - \frac{1}{v} p u t t i n g t h e v a l u e o f v w e g e t 4 l o g (x \times \frac{y}{x}) + l o g c = - \frac{x}{y} \Rightarrow 4 l o g (y) + l o g c = - \frac{x}{y} \Rightarrow \log (y^{4} c) = - \frac{x}{y} \Rightarrow y^{4} c = e^{- \frac{x}{y}}$

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Question 17:

Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e^2x.

Answer:

$According to the question, \frac{d y}{d x} = y + 2 x \Rightarrow \frac{d y}{d x} - y = 2 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - 1 and Q = 2 x ∴ I . F . = e^{\int P d x} = e^{- \int d x} = e^{- x} Multiplying both sides of (1) by I . F . = e^{- x}, we get e^{- x} (\frac{d y}{d x} - y) = e^{- x} 2 x \Rightarrow e^{- x} \frac{d y}{d x} - e^{- x} y = e^{- x} 2 x Integrating both sides with respect to x, we get y e^{- x} = 2 \int \underset{II}{e^{- x}} \underset{I}{x} d x + C \Rightarrow y e^{- x} = 2 x \int e^{- x} d x - 2 \int [\frac{d}{d x} (x) \int e^{- x} d x] d x + C \Rightarrow y e^{- x} = - 2 x e^{- x} - 2 e^{- x} + C . . . . . (2) Since the curve passes through origin, we have 0 \times e^{0} = - 2 \times 0 \times e^{0} - 2 e^{0} + C \Rightarrow C = 2 Putting the value of C in (2), we get y e^{- x} = - 2 x e^{- x} - 2 e^{- x} + 2 \Rightarrow y = - 2 x - 2 + 2 e^{x} \Rightarrow y + 2 (x + 1) = 2 e^{x} DISCLAIMER : In the question it should be e^{x} instead of e^{2 x} .$

Page No 22.135:

Question 18:

The tangent at any point (x, y) of a curve makes an angle tan⁻¹(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).

Answer:

The slope of the curve is given as $\frac{d y}{d x} = \tan θ$ .
Here,

$θ = \tan^{- 1} (2 x + 3 y) ∴ \frac{d y}{d x} = \tan (\tan^{- 1} 2 x + 3 y) \Rightarrow \frac{d y}{d x} = 2 x + 3 y$

$\Rightarrow \frac{d y}{d x} - 3 y = 2 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - 3 and Q = 2 x ∴ I . F . = e^{\int P d x} = e^{\int - 3 d x} = e^{- 3 x} Multiplying both sides of (1), by I . F . = e^{- 3 x}, we get e^{- 3 x} (\frac{d y}{d x} - 3 y) = e^{- 3 x} . 2 x \Rightarrow e^{- 3 x} (\frac{d y}{d x} - 3 y) = 2 x e^{- 3 x} Integrating both sides with respect to x, we get y e^{- 3 x} = 2 \int x e^{- 3 x} d x + C \Rightarrow y e^{- 3 x} = 2 \int \underset{I}{x} \underset{II}{e^{- 3 x}} d x + C \Rightarrow y e^{- 3 x} = 2 x \int e^{- 3 x} d x - 2 \int [\frac{d}{d x} (x) \int e^{- 3 x} d x] d x + C \Rightarrow y e^{- 3 x} = - 2 x \frac{e^{- 3 x}}{3} + 2 \times \frac{1}{3} \int e^{- 3 x} d x + C \Rightarrow y e^{- 3 x} = - \frac{2}{3} x e^{- 3 x} - 2 \times \frac{1}{9} e^{- 3 x} + C \Rightarrow y e^{- 3 x} = - \frac{2}{3} x e^{- 3 x} - \frac{2}{9} e^{- 3 x} + C Since the curve passes through (1, 2), it satisfies the above equation . ∴ 2 e^{- 3} = - \frac{2}{3} e^{- 3} - \frac{2}{9} e^{- 3} + C \Rightarrow C = 2 e^{- 3} + \frac{2}{3} e^{- 3} + \frac{2}{9} e^{- 3} \Rightarrow C = \frac{26}{9} e^{- 3} Putting the value of C, we get y e^{- 3 x} = (- \frac{2}{3} x - \frac{2}{9}) e^{- 3 x} + \frac{26}{9} e^{- 3}$

Page No 22.135:

Question 19:

Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).

Answer:

Portion of the x-axis cut off between the origin and tangent at a point $= x - y \frac{d x}{d y} = OT$

It is given, OT = 2x

$\begin{array}{l} ∴ x - y \frac{d x}{d y} = 2 x \\ - x = y \frac{d x}{d y} \\ - \int \frac{d x}{x} = \int \frac{d y}{y} \\ ∴ x y = k \end{array}$

Since the curve passes through the point (1, 2)
⇒ at x = 1 ⇒ y = 2
∴ k = 2
∴ xy = 2

Page No 22.135:

Question 20:

Find the equation to the curve satisfying x (x + 1) $\frac{d y}{d x} - y$ = x (x + 1) and passing through (1, 0).

Answer:

$We have, x (x + 1) \frac{d y}{d x} - y = x (x + 1) \Rightarrow \frac{d y}{d x} - \frac{y}{x (x + 1)} = 1 Comparing with \frac{d y}{d x} + P y = Q, we get P = - \frac{1}{x (x + 1)} Q = 1 Now, I . F . = e^{- \int \frac{1}{x (x + 1)} d x} = e^{- \int \frac{1}{x} - \frac{1}{x + 1} d x} = e^{- \log |\frac{x}{x + 1}|} = \frac{x + 1}{x} So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow (\frac{x + 1}{x}) y = \int \frac{x + 1}{x} d x + C \Rightarrow (\frac{x + 1}{x}) y = \int d x + \int \frac{1}{x} d x + C \Rightarrow (\frac{x + 1}{x}) y = x + \log |x| + C Since the curve passes throught the point (1, 0), it satisfies the equation of the curve . \Rightarrow (\frac{1 + 1}{1}) 0 = 1 + \log |1| + C \Rightarrow C = - 1 Putting the value of C in the equation of the curve, we get (\frac{x + 1}{x}) y = x + \log |x| - 1 \Rightarrow y = \frac{x}{x + 1} (x + \log |x| - 1)$

Page No 22.135:

Question 21:

Find the equation of the curve which passes through the point (3, −4) and has the slope $\frac{2 y}{x}$ at any point (x, y) on it.

Answer:

According to the question,
$\frac{d y}{d x} = \frac{2 y}{x}$
$\Rightarrow \frac{1}{2 y} d y = \frac{1}{x} d x Integrating both sides with respect to x, we get \int \frac{1}{2 y} d y = \int \frac{1}{x} d x \Rightarrow \frac{1}{2} \log |y| = \log |x| + C Since the curve passes through (3, - 4), it satisfies the above equation . ∴ \frac{1}{2} \log |- 4| = \log |3| + C \Rightarrow \log |2| - \log |3| = C \Rightarrow C = \log |\frac{2}{3}| Putting the value of C, we get \log |y| = 2 \log |x| + 2 \log |\frac{2}{3}| \Rightarrow \log |y| = \log |\frac{4}{9} x^{2}| \Rightarrow y = \pm \frac{4}{9} x^{2} \Rightarrow 9 y - 4 x^{2} = 0 o r 9 y + 4 x^{2} = 0 The given point does not satisfy the equation 9 y - 4 x^{2} = 0 . ∴ 9 y + 4 x^{2} = 0$

Page No 22.135:

Question 22:

Find the equation of the curve which passes through the origin and has the slope x + 3y − 1 at any point (x, y) on it.

Answer:

According to the question,
$\frac{d y}{d x} = x + 3 y - 1$

$\Rightarrow \frac{d y}{d x} - 3 y = x - 1$
$Comparing with \frac{d y}{d x} + P y = Q, we get P = - 3 Q = x - 1 Now, I . F . = e^{- \int 3 d x} = e^{- 3 x} So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow y e^{- 3 x} = \int (x - 1) e^{- 3 x} d x + C \Rightarrow y e^{- 3 x} = \int \underset{I}{x} \underset{II}{e^{- 3 x}} d x - \int e^{- 3 x} d x + C \Rightarrow y e^{- 3 x} = x \int e^{- 3 x} d x - \int [\frac{d}{d x} (x) \int e^{- 3 x} d x] d x - \int e^{- 3 x} d x + C \Rightarrow y e^{- 3 x} = - \frac{1}{3} x e^{- 3 x} + \frac{1}{3} \int e^{- 3 x} d x - \int e^{- 3 x} d x + C \Rightarrow y e^{- 3 x} = - \frac{1}{3} x e^{- 3 x} - \frac{1}{9} e^{- 3 x} + \frac{1}{3} e^{- 3 x} + C \Rightarrow y = - \frac{1}{3} x - \frac{1}{9} + \frac{1}{3} + C e^{3 x} \Rightarrow y = - \frac{1}{3} x + \frac{2}{9} + C e^{3 x} Since the curve passes throught the origin, it satisfies the equation of the curve . \Rightarrow 0 = - 0 + \frac{2}{9} + C e^{0} \Rightarrow C = - \frac{2}{9} Putting the value of C in the equation of the curve, we get y = - \frac{1}{3} x + \frac{2}{9} (1 - e^{3 x}) \Rightarrow y + \frac{1}{3} x = \frac{2}{9} (1 - e^{3 x}) \Rightarrow 3 (3 y + x) = 2 (1 - e^{3 x})$

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Question 23:

At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.

Answer:

According to the question,

$\frac{d y}{d x} = x + x y \Rightarrow \frac{d y}{d x} = x (1 + y)$

$\Rightarrow \frac{1}{1 + y} d y = x d x Integrating both sides with respect to x, we get \int \frac{1}{1 + y} d y = \int x d x \Rightarrow \log |1 + y| = \frac{x^{2}}{2} + C Since the curve passes through (0, 1), it satisfies the above equation . ∴ \log |1 + 1| = \frac{0}{2} + C \Rightarrow C = \log 2 Putting the value of C, we get \log |1 + y| = \frac{x^{2}}{2} + \log 2 \Rightarrow \log |\frac{1 + y}{2}| = \frac{x^{2}}{2} \Rightarrow \frac{1 + y}{2} = e^{\frac{x^{2}}{2}} \Rightarrow y + 1 = 2 e^{\frac{x^{2}}{2}}$

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Question 24:

A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is $y^{2} - 2 x y \frac{d y}{d x} - x^{2} = 0$ , and hence find the curve.

Answer:

Tangent at P(x, y) is given by $Y - y = \frac{d y}{d x} (X - x)$
If p be the perpendicular from the origin, then
$p = \frac{x \frac{d y}{d x} - y}{\sqrt{[1 + {(\frac{d y}{d x})}^{2}]}} = x (g i v e n) \Rightarrow x^{2} {(\frac{d y}{d x})}^{2} - 2 x y \frac{d y}{d x} + y^{2} = x^{2} + x^{2} {(\frac{d y}{d x})}^{2} \Rightarrow y^{2} - 2 x y \frac{d y}{d x} - x^{2} = 0 H e n c e p r o v e d . N o w, y^{2} - 2 x y \frac{d y}{d x} - x^{2} = 0 \Rightarrow \frac{d y}{d x} = \frac{y^{2} - x^{2}}{2 x y} \Rightarrow 2 x y \frac{d y}{d x} - y^{2} = - x^{2} \Rightarrow 2 y \frac{d y}{d x} - \frac{y^{2}}{x} = - x^{} L e t y^{2} = v \Rightarrow \frac{d v}{d x} - \frac{v}{x} = - x$
Multiplying by the integrating factor $e^{\int - \frac{1}{x} d x} = \frac{1}{x}$
$v . \frac{1}{x} = \int - x . \frac{1}{x} d x + c = - x + c \Rightarrow \frac{y^{2}}{x^{2}} = - x + c \Rightarrow x^{2} + y^{2} = c x$

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Question 25:

Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.

Answer:

It is given that the distance between the foot of ordinate of point of contact (A) and point of intersection of tangent with x-axis (T) = 2x

$Coordinate of T = (x - y \frac{d x}{d y}, 0) A T = [x - (x - y \frac{d x}{d y})] = 2 x [Equation of tangent, Y - y = \frac{d y}{d x} (X - x)] \Rightarrow y - 0 = \frac{d y}{d x} (x - (x - y \frac{d x}{d y})) \Rightarrow y \frac{d x}{d y} = 2 x \Rightarrow \int \frac{d x}{x} = 2 \int \frac{d y}{y} \Rightarrow \ln x = \ln y^{2} + \ln c x = c y^{2} As the circle passes through (1, 2) . 1 = c \times 2^{2} \Rightarrow c = \frac{1}{4} \Rightarrow 4 x = y^{2}$

Page No 22.135:

Question 26:

The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.

Answer:

Let P (x, y) be any point on the curve. The equation of the normal at P (x, y) to the given curve is given as
$Y - y = - \frac{1}{\frac{d y}{d x}} (X - x)$
It is given that the curve passes through the point (3, 0). Then,
$0 - y = - \frac{1}{\frac{d y}{d x}} (3 - x) \Rightarrow - y = - \frac{1}{\frac{d y}{d x}} (3 - x) \Rightarrow y \frac{d y}{d x} = 3 - x \Rightarrow y d y = (3 - x) d x \Rightarrow \frac{y^{2}}{2} = 3 x - \frac{x^{2}}{2} + C . . . . . (1) Since the curve passes through the point (3, 4), it satisfies the equation . \Rightarrow \frac{4^{2}}{2} = 3 (3) - \frac{3^{2}}{2} + C \Rightarrow C = 8 - 9 + \frac{9}{2} \Rightarrow C = \frac{9}{2} - 1 = \frac{7}{2} Putting the value of C in (1), we get \frac{y^{2}}{2} = 3 x - \frac{x^{2}}{2} + \frac{7}{2} \Rightarrow y^{2} = 6 x - x^{2} + 7 \Rightarrow x^{2} + y^{2} - 6 x - 7 = 0$

Page No 22.135:

Question 27:

The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.

Answer:

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: $\frac{d P}{d t} α P$
$\Rightarrow \frac{d P}{d t} = a P \Rightarrow \frac{d P}{P} = a d t \Rightarrow \log |P| = a t + C . . . . . (1) Now, P = N at t = 0 Putting P = N and t = 0 in (1), we get \log |N| = C Putting C = \log |N| in (1), we get \log |P| = a t + \log |N| \Rightarrow \log |\frac{P}{N}| = a t . . . . . (2) According to the question, \log |\frac{2 N}{N}| = 6 a \Rightarrow a = \frac{1}{6} \log |2| Putting a = \frac{1}{6} \log |2| in (2), we get \log |\frac{P}{N}| = \frac{t}{6} \log |2| . . . . . (3) Putting t = 18 in (3) to find the bacteria after 18 hours, we get \log |\frac{P}{N}| = \frac{18}{6} \log |2| \Rightarrow \log |\frac{P}{N}| = 3 \log |2| \Rightarrow \log |\frac{P}{N}| = \log |8| \Rightarrow \frac{P}{N} = 8 \Rightarrow P = 8 N$

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Question 28:

Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, approximately 1.1% of a certain quantity of radium has decomposed. Determine approximately how long it will take for one-half of the original amount of radium to decompose?

Answer:

Let the original amount of radium be N and the amount of radium at any time t be P.
Given: $\frac{d P}{d t} α P$
$\Rightarrow \frac{d P}{d t} = - a P \Rightarrow \frac{d P}{P} = - a d t Integrating both sides, we get \Rightarrow \log |P| = - a t + C . . . . . (1) Now, P = N when t = 0 Putting P = N and t = 0 in (1), we get \log |N| = C Putting C = \log |N| in (1), we get \log |P| = - a t + \log |N| \Rightarrow \log |\frac{P}{N}| = - a t . . . . . (2) According to the question, P = \frac{98.9}{100} N = 0.989 N a t t = 25 ∴ \log |\frac{0.989 N}{N}| = - 25 a \Rightarrow a = - \frac{1}{25} \log |0.989| Putting a = - \frac{1}{25} \log |0.989| in (2), we get \log |\frac{P}{N}| = (\frac{1}{25} \log |0.989|) t To find the time when the radium becomes half of its quantity, we have N = \frac{1}{2} P ∴ \log |\frac{N}{\frac{N}{2}}| = (\frac{1}{25} \log |0.989|) t \Rightarrow \log |2| = (\frac{1}{25} \log |0.989|) t \Rightarrow t = \frac{25 \log 2}{\log 0.989} = \frac{25 \times 0.6931}{0.01106} = 1566.68 \approx 1567 (approx .)$

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Question 29:

Show that all curves for which the slope at any point (x, y) on it is $\frac{x^{2} + y^{2}}{2 x y}$ are rectangular hyperbola.

Answer:

We have,
$\frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 x y} Let y = v x \frac{d y}{d x} = v + x \frac{d v}{d x} ∴ v + x \frac{d v}{d x} = \frac{x^{2} + v^{2} x^{2}}{2 v x^{2}} \Rightarrow x \frac{d v}{d x} = \frac{1 + v^{2}}{2 v} - v \Rightarrow \frac{x d v}{d x} = \frac{1 + v^{2} - 2 v^{2}}{2 v} \Rightarrow x \frac{d v}{d x} = \frac{1 - v^{2}}{2 v} \Rightarrow \frac{2 v}{1 - v^{2}} d v = \frac{1}{x} d x$

$Integrating both sides, we get \int \frac{2 v}{1 - v^{2}} d y = \int \frac{1}{x} d x \Rightarrow - \log |1 - v^{2}| = \log |x| - \log |C| \Rightarrow \log |\frac{1 - v^{2}}{C}| = - \log |x| \Rightarrow 1 - v^{2} = \frac{C}{x} \Rightarrow 1 - {(\frac{y}{x})}^{2} = \frac{C}{x} \Rightarrow \frac{x^{2} - y^{2}}{x^{2}} = \frac{C}{x} \Rightarrow x^{2} - y^{2} = C x$
Thus, $x^{2} - y^{2} = C x$ is the equation of the rectangular hyperbola.

Page No 22.136:

Question 30:

The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.

Answer:

According to the question,
$\frac{d y}{d x} = x + y$
$\Rightarrow \frac{d y}{d x} - y = x$
$Comparing with \frac{d y}{d x} + P y = Q, we get P = - 1 Q = x Now, I . F . = e^{- \int d x} = e^{- x} So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow y e^{- x} = \int \underset{I}{x} \underset{II}{e^{- x}} d x + C \Rightarrow y e^{- x} = x \int e^{- x} d x - \int [\frac{d}{d x} (x) \int e^{- x} d x] d x + C \Rightarrow y e^{- x} = - x e^{- x} + \int e^{- x} d x + C \Rightarrow y e^{- x} = - x e^{- x} - e^{- x} + C Since the curve passes throught the origin, it satisfies the equation of the curve . \Rightarrow 0 e^{0} = - 0 e^{0} - e^{0} + C \Rightarrow C = 1 Putting the value of C in the equation of the curve, we get y e^{- x} = - x e^{- x} - e^{- x} + 1 \Rightarrow y e^{- x} + x e^{- x} + e^{- x} = 1 \Rightarrow (y + x + 1) e^{- x} = 1 \Rightarrow (x + y + 1) = e^{x}$

Page No 22.136:

Question 31:

Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.

Answer:

According to the question,
$\frac{d y}{d x} = x + x y$
$\Rightarrow \frac{d y}{d x} - x y = x$
$Comparing with \frac{d y}{d x} + P y = Q, we get P = - x Q = x Now, I . F . = e^{- \int x d x} = e^{- \frac{x^{2}}{2}} So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow y e^{- \frac{x^{2}}{2}} = \int x e^{- \frac{x^{2}}{2}} d x + C \Rightarrow y e^{- \frac{x^{2}}{2}} = I + C Now, I = \int x e^{- \frac{x^{2}}{2}} d x Putting \frac{- x^{2}}{2} = t, we get - x d x = d t ∴ I = - \int e^{t} d t \Rightarrow I = - e^{t} \Rightarrow I = - e^{\frac{- x^{2}}{2}} ∴ y e^{- \frac{x^{2}}{2}} = - e^{\frac{- x^{2}}{2}} + C Since the curve passes throught the point (0, 1), it satisfies the equation of the curve . \Rightarrow 1 e^{0} = - e^{0} + C \Rightarrow C = 2 Putting the value of C in the equation of the curve, we get y e^{- \frac{x^{2}}{2}} = - e^{\frac{- x^{2}}{2}} + 2 \Rightarrow y = - 1 + 2 e^{\frac{x^{2}}{2}}$

Page No 22.136:

Question 32:

The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).

Answer:

According to the question,
$\frac{d y}{d x} = x^{2}$
$\Rightarrow d y = x^{2} d x Integrating both sides with respect to x, we get \int d y = \int x^{2} d x \Rightarrow y = \frac{x^{3}}{3} + C Since the curve passes through (- 1, 1), it satisfies the above equation . ∴ 1 = \frac{- 1}{3} + C \Rightarrow C = 1 + \frac{1}{3} \Rightarrow C = \frac{4}{3} Putting the value of C, we get y = \frac{x^{3}}{3} + \frac{4}{3} \Rightarrow 3 y = x^{3} + 4$

Page No 22.136:

Question 33:

Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.

Answer:

According to the question,

$y \frac{d y}{d x} = x \Rightarrow y d y = x d x Integrating both sides with respect to x, we get \int y d y = \int x d x \Rightarrow \frac{y^{2}}{2} = \frac{x^{2}}{2} + C Since the curve passes through (0, a), it satisfies the above equation . ∴ \frac{a^{2}}{2} = \frac{0}{2} + C \Rightarrow C = \frac{a^{2}}{2} Putting the value of C, we get \frac{y^{2}}{2} = \frac{x^{2}}{2} + \frac{a^{2}}{2} \Rightarrow x^{2} - y^{2} = - a^{2}$

Page No 22.136:

Question 34:

The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).

Answer:

Let P(x, y) be any point on the curve. Then slope of the tangent at P is $\frac{d y}{d x}$ .
It is given that the slope of the tangent at P(x,y) is equal to the ordinate i.e y.
Therefore $\frac{d y}{d x}$ = y
$\Rightarrow \frac{1}{y} d y = d x \Rightarrow \log y = x + \log C \Rightarrow l o g y = \log e^{x} + l o g C \Rightarrow y = C e^{x}$
Since, the curve passes through (1,1). Therefore, x=1 and y=1 .
Putting these values in equation obtained above we get,
$1 = C e^{1} \Rightarrow C = \frac{1}{e} p u t t i n g t h e s e v a l u e s i n t h e e q u a t i o n w e g e t, y = e^{x - 1}$

Page No 22.137:

Question 1:

Define a differential equation.

Answer:

Differential equation:

An equation containing an independent variable, a dependent variable and differential coefficients of the dependent variable with respect to the independent variable is called a differential equation.
for example: $\frac{d y}{d x} = e^{x + y}$

Page No 22.137:

Question 2:

Define order of a differential equation.

Answer:

Order of differential equation:

The order of a differential equation is the order of its highest order derivative that apears in the equation.

example: $\frac{d^{2} y}{d x^{2}} - 4 (\frac{d y}{d x}) = 2 y$
order of the differential equation is 2.

Page No 22.137:

Question 3:

Define degree of a differential equation.

Answer:

Degree of differential equation:

The degree of a differential equation is the power of the highest order derivative occurring in a differential equation when it is written as a polynomial in differential coefficients.

example: ${(\frac{d^{2} y}{d x^{2}})}^{2} - 4 (\frac{d y}{d x}) = 2 y$
the degree of the given differential equation is 2

Page No 22.137:

Question 4:

Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.

Answer:

$We have, y = C x + 5 . . . . . (1) \Rightarrow \frac{d y}{d x} = C Substituting the value of C in (1), we get y = \frac{d y}{d x} \times x + 5 \Rightarrow x \frac{d y}{d x} - y + 5 = 0 Hence, x \frac{d y}{d x} - y + 5 = 0 is the differential equation representing the family of straight lines y = C x + 5, where C is an arbitary constant .$

Page No 22.137:

Question 5:

Write the differential equation obtained by eliminating the arbitrary constant C in the equation x² − y² = C².

Answer:

$We have, x^{2} - y^{2} = C^{2} Differentiating with respect to x, we get 2 x - 2 y \frac{d y}{d x} = 0 \Rightarrow 2 x = 2 y \frac{d y}{d x} \Rightarrow x d x = y d y \Rightarrow x d x - y d y = 0 Hence, x d x - y d y = 0 is the required differential equation .$

Page No 22.138:

Question 6:

Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C².

Answer:

$We have, x y = C^{2} Differentiating with respect to x, we get x \frac{d y}{d x} + y = 0 \Rightarrow x \frac{d y}{d x} = - y \Rightarrow x d y = - y d x \Rightarrow x d y + y d x = 0 Hence, x d y + y d x = 0 is the required differential equation .$

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Question 7:

Write the degree of the differential equation $a^{2} \frac{d^{2} y}{d x^{2}} = {\{1 + {(\frac{d y}{d x})}^{2}\}}^{1 / 4}$ .

Answer:

$We have, a^{2} \frac{d^{2} y}{d x^{2}} = {\{1 + {(\frac{d y}{d x})}^{2}\}}^{1 / 4} {\{a^{2} \frac{d^{2} y}{d x^{2}}\}}^{4} = 1 + {(\frac{d y}{d x})}^{2} Degree of the differential equation is the degree of the highest order derivative . Therefore, the degree must be 4 .$

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Question 8:

Write the order of the differential equation $1 + {(\frac{d y}{d x})}^{2} = 7 {(\frac{d^{2} y}{d x^{2}})}^{3}$ .

Answer:

$1 + {(\frac{d y}{d x})}^{2} = 7 {(\frac{d^{2} y}{d x^{2}})}^{3} The order of a differential equation is the order of its highest order derivatives . Here, the required order is 2 .$

Page No 22.138:

Question 9:

Write the order and degree of the differential equation $y = x \frac{d y}{d x} + a \sqrt{1 + {(\frac{d y}{d x})}^{2}}$ .

Answer:

$We have, y = x \frac{d y}{d x} + a \sqrt{1 + {(\frac{d y}{d x})}^{2}} y - x \frac{d y}{d x} = a \sqrt{1 + {(\frac{d y}{d x})}^{2}} Squaring both sides, we get {(y - x \frac{d y}{d x})}^{2} = {\{a \sqrt{1 + {(\frac{d y}{d x})}^{2}}\}}^{2} y^{2} + x^{2} {(\frac{d y}{d x})}^{2} - 2 x y \frac{d y}{d x} = a^{2} \{1 + {(\frac{d y}{d x})}^{2}\} y^{2} + {(\frac{d y}{d x})}^{2} \{x^{2} - a^{2}\} - 2 x y \frac{d y}{d x} = a^{2} From the above equation, we see that the highest order is 1 . So, its order is 1 and the power of the highest order derivative is 2 . Thus, it is a differential equation of order 1 and degree 2 .$

Page No 22.138:

Question 10:

Write the degree of the differential equation $\frac{d^{2} y}{d x^{2}} + x {(\frac{d y}{d x})}^{2} = 2 x^{2} \log (\frac{d^{2} y}{d x^{2}})$ .

Answer:

$We have, \frac{d^{2} y}{d x^{2}} + x {(\frac{d y}{d x})}^{2} = 2 x^{2} \log (\frac{d^{2} y}{d x^{2}}) \Rightarrow \frac{d^{2} y}{d x^{2}} + x {(\frac{d y}{d x})}^{2} - 2 x^{2} \log (\frac{d^{2} y}{d x^{2}}) = 0 Here, we observe that LHS of the differential equation cannot be expressed as a polynomial in \frac{d y}{d x} . So, its degree is not defined .$

Page No 22.138:

Question 11:

Write the order of the differential equation of the family of circles touching X-axis at the origin.

Answer:

The equation of the family of circles touching x-axis at the origin is
${(x - 0)}^{2} + {(y - a)}^{2} = a^{2} x^{2} + y^{2} - 2 a y = 0 . . . . . (1) Here, a is the parameter . Since this equation contains only one arbitary constant, we differentiate it only once . Differentiating with respect to x, we get 2 x + 2 y \frac{d y}{d x} - 2 a \frac{d y}{d x} = 0 a = \frac{x + y (\frac{d y}{d x})}{\frac{d y}{d x}} . . . . . (2) Putting the value of a from (2) in (1), we get x^{2} + y^{2} = 2 y \{\frac{x + y (\frac{d y}{d x})}{\frac{d y}{d x}}\} (x^{2} - y^{2}) \frac{d y}{d x} = 2 x y So, this is the required differential equation . Here, order of the differential equation is 1 .$

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Question 12:

Write the order of the differential equation of all non-horizontal lines in a plane.

Answer:

$The equation of the non - horizontal lines in a plane is y = m x + c, where m is the slope and c is the intercept on y - axis . Differentiating with respect to x, we get \frac{d y}{d x} = m \Rightarrow \frac{d^{2} y}{d x^{2}} = 0 This is the required differential equation . Here, we observe that the order of the required differential equation is 2 .$

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Question 13:

If sin x is an integrating factor of the differential equation $\frac{d y}{d x} + P y = Q$ , then write the value of P.

Answer:

$It is given that \sin x is the integrating factor of the differential equation \frac{d y}{d x} + P y = Q . ∴ e^{\int P d x} = \sin x \Rightarrow \int P d x = \log |\sin x| \Rightarrow \int P d x = \int \cot x d x [∵ \int \cot x d x = \log |\sin x| + C] \Rightarrow P = \cot x$

Page No 22.138:

Question 14:

Write the order of the differential equation of the family of circles of radius r.

Answer:

ans

Page No 22.138:

Question 15:

Write the order of the differential equation whose solution is y = a cos x + b sin x + c e^−x.

Answer:

$y = a \cos x + b \sin x + c e^{- x} Here, we see that there are three arbitary constants . Therefore, we differentiate it three times to get rid of all three arbitrary constants . Hence, the order of the differential equation is 3 .$

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Question 16:

Write the order of the differential equation associated with the primitive y = C₁ + C₂e^x + C₃e^−2x^{+ C₄}, where C₁, C₂, C₃, C₄ are arbitrary constants.

Answer:

$y = C_{1} + C_{2} e^{x} + C_{3} e^{- 2 x + C_{4}} the given equation can be reduced to : y = C_{1} + C_{2} e^{x} + C_{3} (e^{- 2 x} \times e^{c_{4}}) Here, e^{c_{4}} w ill be a constant . W e h a v e 3 c o n s \tan t s a s C_{1}, C_{2} a n d C_{3} . and a differential equation of order n always contains exactly n essential arbitrary constants . Hence, the order of the required differntial equation is 3 .$

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Question 17:

What is the degree of the following differential equation?
$5 x {(\frac{d y}{d x})}^{2} - \frac{d^{2} y}{d x^{2}} - 6 y = \log x$

Answer:

$5 x {(\frac{d y}{d x})}^{2} - \frac{d^{2} y}{d x^{2}} - 6 y = \log x$
Here, we see that the highest order derivative is $\frac{d^{2} y}{d x^{2}}$ and its power is 1.
Therefore, the given differential equation is of first degree.

Page No 22.138:

Question 18:

Write the degree of the differential equation ${(\frac{d y}{d x})}^{4} + 3 x \frac{d^{2} y}{d x^{2}} = 0$ .

Answer:

${(\frac{d y}{d x})}^{4} + 3 x \frac{d^{2} y}{d x^{2}} = 0$
The highest order derivative is $\frac{d^{2} y}{d x^{2}}$ and its power is 1.
Therefore, the given differential equation is of first degree.

Page No 22.138:

Question 19:

Write the degree of the differential equation x ${(\frac{d^{2} y}{d x^{2}})}^{3} + y {(\frac{d y}{d x})}^{4} + x^{3} = 0$ .

Answer:

${(\frac{d^{2} y}{d x^{2}})}^{3} + y {(\frac{d y}{d x})}^{4} + x^{3} = 0$

The highest order derivative is $\frac{d^{2} y}{d x^{2}}$ and its power is 3.

Therefore, the degree of given differential equation is 3.

Page No 22.138:

Question 20:

Write the differential equation representing family of curves y = mx, where m is arbitrary constant.

Answer:

$We have, y = m x . . . . . (1) Differentiating with respect to x \Rightarrow \frac{d y}{d x} = m Substituting the value of m = \frac{d y}{d x} in eq (1) we get, y = x \frac{d y}{d x} Hence, y = x \frac{d y}{d x} is the required differential equation .$

Page No 22.138:

Question 21:

Write the degree of the differential equation $x^{3} {(\frac{d^{2} y}{d x^{2}})}^{2} + x {(\frac{d y}{d x})}^{4} = 0$ .

Answer:

$x^{3} {(\frac{d^{2} y}{d x^{2}})}^{2} + x {(\frac{d y}{d x})}^{4} = 0$

Here, the highest order derivative is $\frac{d^{2} y}{d x^{2}}$ and its power is 2.

Therefore, degree of given differential equation is 2.

Page No 22.138:

Question 22:

Write the degree of the differential equation ${(1 + \frac{d y}{d x})}^{3} = {(\frac{d^{2} y}{d x^{2}})}^{2}$

Answer:

The degree is 2 as the highest derivative is of order 2.

Page No 22.138:

Question 23:

Write the degree of the differential equation
$\frac{d^{2} y}{d x^{2}} + 3 {(\frac{d y}{d x})}^{2} = x^{2} \log (\frac{d^{2} y}{d x^{2}})$

Answer:

The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 22.139:

Question 24:

Write the degree of the differential equation
${(\frac{d^{2} y}{d x^{2}})}^{2} + {(\frac{d y}{d x})}^{2} = x \sin (\frac{d y}{d x})$

Answer:

The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 22.139:

Question 25:

Write the order and degree of the differential equation $\frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{\frac{1}{4}} + x^{\frac{1}{5}} = 0$

Answer:

The order is 2 as the highest derivative is 2.
The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 22.139:

Question 26:

The degree ofthe differential equation $\frac{d^{2} y}{d x^{2}} + e^{\frac{d y}{d x}} = 0$

Answer:

The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 22.139:

Question 27:

How many arbitrary constants are there in the general solution of the differential equation of order 3.

Answer:

The arbitrary constants in the general solution of the differential equation is equal to the order of the differential equation.
Hence, the number of arbitrary constants in the general solution of the differential equation of order 3 are 3.

Page No 22.139:

Question 28:

Write the order of the differential equation representing the family of curves y = ax + a³.

Answer:

The order of the differential equation is equal to the arbitrary constants present in the general solution of the differential equation.
Hence, the order of the differential equation representing the family of curves y = ax + a³ is 1.

Page No 22.139:

Question 29:

Find the sum of the order and degree of the differential equation $y = x {(\frac{d y}{d x})}^{3} + \frac{d^{2} y}{d x^{2}}$

Answer:

The order is 2 as the highest derivative is 2.
The degree is 1 as the highest derivative is of order 1.
Hence, the sum of the order and degree of the differential equation $y = x {(\frac{d y}{d x})}^{3} + \frac{d^{2} y}{d x^{2}}$ is 2 + 1 = 3

Page No 22.139:

Question 30:

Find the solution of the differential equation
$x \sqrt{1 + y^{2}} d x + y \sqrt{1 + x^{2}} d y = 0$

Answer:

$x \sqrt{1 + y^{2}} d x + y \sqrt{1 + x^{2}} d y = 0 \Rightarrow y \sqrt{1 + x^{2}} d y = - x \sqrt{1 + y^{2}} d x \Rightarrow \frac{y}{\sqrt{1 + y^{2}}} d y = \frac{- x}{\sqrt{1 + x^{2}}} d x \Rightarrow \int \frac{y}{\sqrt{1 + y^{2}}} d y = - \int \frac{x}{\sqrt{1 + x^{2}}} d x$
$Let 1 + y^{2} = t^{2} and 1 + x^{2} = p^{2} \Rightarrow 2 y d y = 2 t d t and 2 x d x = 2 p d p \Rightarrow y d y = t d t and x d x = p d p Substituting in above equation, we get \Rightarrow \int d t = - \int d p \Rightarrow t = - p + C \Rightarrow \sqrt{1 + x^{2}} + \sqrt{1 + y^{2}} = C$

Page No 22.139:

Question 1:

The integrating factor of the differential equation (x log x) $\frac{d y}{d x} + y = 2 \log x$ , is given by
(a) log (log x)

(b) e^x

(c) log x

(d) x

Answer:

(c) log x

We have,
(x log x) $\frac{d y}{d x} + y = 2 \log x$
Dividing both sides by x log x, we get
$\frac{d y}{d x} + \frac{y}{x \log x} = 2 \frac{\log x}{x \log x} \Rightarrow \frac{d y}{d x} + \frac{y}{x \log x} = \frac{2}{x} \Rightarrow \frac{d y}{d x} + (\frac{1}{x \log x}) y = \frac{2}{x} Comparing with \frac{d y}{d x} + P y = Q, we get P = \frac{1}{x \log x} Q = \frac{2}{x} Now, I . F . = e^{\int P d x} = e^{\int \frac{1}{x \log x} d x} = e^{\log (\log x)} = \log x$

Page No 22.139:

Question 2:

The general solution of the differential equation $\frac{d y}{d x} = \frac{y}{x}$ is
(a) log y = kx

(b) y = kx

(c) xy = k

(d) y = k log x

Answer:

(b) y = kx

We have,
$\frac{d y}{d x} = \frac{y}{x} \Rightarrow \frac{1}{y} d y = \frac{1}{x} d x Integrating both sides, we get \int \frac{1}{y} d y = \int \frac{1}{x} d x \Rightarrow \log y = \log x + \log k \Rightarrow \log y - \log x = \log k \Rightarrow \log (\frac{y}{x}) = \log k \Rightarrow \frac{y}{x} = k \Rightarrow y = k x$

Page No 22.139:

Question 3:

Integrating factor of the differential equation cos $x \frac{d y}{d x} + y$ sin x = 1, is
(a) sin x
(b) sec x
(c) tan x
(d) cos x

Answer:

(b) sec x

We have,
$\cos x \frac{d y}{d x} + y \sin x = 1$
Dividing both sides by cos x, we get
$\frac{d y}{d x} + \frac{\sin x}{\cos x} y = \frac{1}{\cos x} \Rightarrow \frac{d y}{d x} + (\tan x) y = \frac{1}{\cos x} Comparing with \frac{d y}{d x} + P y = Q, we get P = \tan x Q = \frac{1}{\cos x} Now, I . F . = e^{\int \tan x d x} = e^{\log (\sec x)} = \sec x$

Page No 22.139:

Question 4:

The degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{2} - (\frac{d y}{d x}) = y^{3}$ , is
(a) 1/2
(b) 2
(c) 3
(d) 4

Answer:

(b) 2

We have,
${(\frac{d^{2} y}{d x^{2}})}^{2} - (\frac{d y}{d x}) = y^{3}$
$The highest order derivative is \frac{d^{2} y}{d^{2} x} and its power is 2 . Hence, the degree is 2 .$

Page No 22.140:

Question 5:

The degree of the differential equation ${\{5 + {(\frac{d y}{d x})}^{2}\}}^{5 / 3} = x^{5} (\frac{d^{2} y}{d x^{2}})$ , is
(a) 4
(b) 2
(c) 5
(d) 10

Answer:

We have,
${[5 + {(\frac{d y}{d x})}^{2}]}^{\frac{5}{3}} = x^{5} (\frac{d^{2} y}{d^{2} x}) Taking Cube power on both sides, we get {(5 + {(\frac{d y}{d x})}^{2})}^{5} = x^{15} {(\frac{d^{2} y}{d^{2} x})}^{3} The highest order derivative is \frac{d^{2} y}{d^{2} x} and its power is 3 . Hence, the degree is 3 .$

Disclaimer: The correct option is not given in the question.

Page No 22.140:

Question 6:

The general solution of the differential equation $\frac{d y}{d x} + y$ cot x = cosec x, is
(a) x + y sin x = C
(b) x + y cos x = C
(c) y + x (sin x + cos x) = C
(d) y sin x = x + C

Answer:

(d) y sin x = x + C

$We have, \frac{d y}{d x} + y \cot x = cosec x \frac{d y}{d x} + y \cot x = cosec x Comparing with \frac{d y}{d x} + P y = Q, we get P = \cot x Q = cosec x Now, I . F . = e^{\int \cot x d x} = e^{\log (\sin x)} = \sin x So, the solution is given by y \sin x = \int \sin x \times cosec x d x + C \Rightarrow y \sin x = x + C$

Page No 22.140:

Question 7:

The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
(a) y" + y' = 0
(b) y" − ω²y = 0
(c) y" = −ω²y
(d) y" + y = 0

Answer:

(c) y" = −ω²y

We have,
y = A cos ωt + B sin ωt .....(1)
Differentiating both sides of (1) with respect to x, we get
$\frac{d y}{d t} = - A ω \sin ω t + B ω \cos ω t$ .....(2)
Differentiating both sides of (2) again with respect to x, we get
$\frac{d^{2} y}{d t^{2}} = - A ω^{2} \cos ω t - B ω^{2} \sin ω t \Rightarrow \frac{d^{2} y}{d t^{2}} = - ω^{2} (A \cos ω t + B \sin ω t) \Rightarrow \frac{d^{2} y}{d t^{2}} = - ω^{2} y [Using (1)] ∴ y'' = - ω^{2} y$

Page No 22.140:

Question 8:

The equation of the curve whose slope is given by $\frac{d y}{d x} = \frac{2 y}{x}; x > 0, y > 0$ and which passes through the point (1, 1) is
(a) x² = y
(b) y² = x
(c) x² = 2y
(d) y² = 2x

Answer:

(a) x² = y

We have,
$\frac{d y}{d x} = \frac{2 y}{x} \Rightarrow \frac{1}{2} \times \frac{1}{y} d y = \frac{1}{x} d x Integrating both sides, we get \frac{1}{2} \int \frac{1}{y} d y = \int \frac{1}{x} d x \Rightarrow \frac{1}{2} \log y = \log x + \log C \Rightarrow \log y^{\frac{1}{2}} - \log x = \log C \Rightarrow \log (\frac{\sqrt{y}}{x}) = \log C \Rightarrow \frac{\sqrt{y}}{x} = C \Rightarrow \sqrt{y} = C x . . . . . (1) As (1) passes through (1, 1), we get ∴ 1 = C Putting the value of C in (1), we get \sqrt{y} = x \Rightarrow y = x^{2}$

Page No 22.140:

Question 9:

The order of the differential equation whose general solution is given by
y = c₁ cos (2x + c₂) − (c₃ + c₄) a^x^{+ c₅} + c₆ sin (x − c₇) is
(a) 3
(b) 4
(c) 5
(d) 2

Answer:

(c) 5

The given equation can be reduced to : $y = c_{1} \cos (2 x + c_{2}) - (c) a^{x} \times a^{c_{5}} + c_{6} \sin (x - c_{7}) where c = c_{3} + c_{4} a n d a^{c_{5}} will be a constant$

There are 5 constants $(c_{1}, c_{2,} c, c_{6}, c_{7})$ in the given differential equation.
Hence, the order of the differential equation is 5.

Page No 22.140:

Question 10:

The solution of the differential equation $\frac{d y}{d x} = \frac{a x + g}{b y + f}$ represents a circle when
(a) a = b
(b) a = −b
(c) a = −2b
(d) a = 2b

Answer:

(b) a = −b

We have,
$\frac{d y}{d x} = \frac{a x + g}{b y + f} \Rightarrow (b y + f) d y = (a x + g) d x Integrating both sides, we get \int (b y + f) d y = \int (a x + g) d x \Rightarrow b \frac{y^{2}}{2} + f y = a \frac{x^{2}}{2} + g x + C \Rightarrow b \frac{y^{2}}{2} + f y - a \frac{x^{2}}{2} - g x = C \Rightarrow b y^{2} + 2 f y - a x^{2} - 2 g x - 2 C = 0 The above equation represents a circle . Therefore, the coffecients of x^{2} and y^{2} must be equal . i . e . - a = b \Rightarrow a = - b$

Page No 22.140:

Question 11:

The solution of the differential equation $\frac{d y}{d x} + \frac{2 y}{x} = 0$ with y(1) = 1 is given by
(a) $y = \frac{1}{x^{2}}$

(b) $x = \frac{1}{y^{2}}$

(c) $x = \frac{1}{y}$

(d) $y = \frac{1}{x}$

Answer:

(a) $y = \frac{1}{x^{2}}$

We have,

$\frac{d y}{d x} + \frac{2 y}{x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2 y}{x} \Rightarrow \frac{1}{2} \times \frac{1}{y} d y = \frac{- 1}{x} d x Integrating both sides, we get \frac{1}{2} \int \frac{1}{y} d y = - \int \frac{1}{x} d x \Rightarrow \frac{1}{2} \log y = - \log x + \log C \Rightarrow \log y^{\frac{1}{2}} + \log x = \log C \Rightarrow \log (\sqrt{y} x) = \log C \Rightarrow \sqrt{y} x = C . . . . . (1) As (1) satisfies y (1) = 1, we get 1 = C Putting the value of C in (1), we get \sqrt{y} x = 1 \Rightarrow y = \frac{1}{x^{2}}$

Page No 22.140:

Question 12:

The solution of the differential equation $\frac{d y}{d x} - \frac{y (x + 1)}{x} = 0$ is given by
(a) y = xe^x^{+ C}
(b) x = ye^x
(c) y = x + C
(d) xy = e^x + C

Answer:

(a) y = xe^x^{+ C}

We have,
$\frac{d y}{d x} - \frac{y (x + 1)}{x} = 0 \Rightarrow \frac{d y}{d x} = \frac{y (x + 1)}{x} \Rightarrow \frac{d y}{y} = \frac{(x + 1)}{x} d x Integrating both sides, we get \int \frac{d y}{y} = \int \frac{(x + 1)}{x} d x \Rightarrow \int \frac{d y}{y} = \int d x + \int \frac{1}{x} d x \Rightarrow \log y = x + \log x + C \Rightarrow \log y - \log x = x + C \Rightarrow \log (\frac{y}{x}) = x + C \Rightarrow \frac{y}{x} = e^{x + C} \Rightarrow y = x e^{x + C}$

Page No 22.140:

Question 13:

The order of the differential equation satisfying $\sqrt{1 - x^{4}} + \sqrt{1 - y^{4}} = a (x^{2} - y^{2})$ is
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(a) 1

The order of a differential equation depends on the number of arbitrary constants in it.

Since $\sqrt{1 - x^{4}} + \sqrt{1 - y^{4}} = a (x^{2} - y^{2})$ contains only 1 constant, the order of the differential equation is 1.

Page No 22.140:

Question 14:

The solution of the differential equation y₁y₃ = y₂² is
(a) x = C₁e^C^₂y + C₃
(b) y = C₁e^C^₂x + C₃
(c) 2x = C₁e^C^₂y + C₃
(d) none of these

Answer:

(b) y = C₁e^C^₂x + C₃

$y_{1} y_{3} = y_{2}^{2} \frac{y_{3}}{y_{2}} = \frac{y_{2}}{y_{1}} \Rightarrow \frac{(\frac{d^{3} y}{d x^{3}})}{(\frac{d^{2} y}{d x^{2}})} = \frac{(\frac{d^{2} y}{d x^{2}})}{(\frac{d y}{d x})} \Rightarrow \int \frac{\frac{d}{d x} (\frac{d^{2} y}{d x^{2}})}{(\frac{d^{2} y}{d x^{2}})} = \int \frac{\frac{d}{d x} (\frac{d y}{d x})}{(\frac{d y}{d x})} \Rightarrow \ln (\frac{d^{2} y}{d x^{2}}) = \ln (\frac{d y}{d x}) + \ln C_{4} \Rightarrow \frac{d^{2} y}{d x^{2}} = C_{4} \frac{d y}{d x} \Rightarrow \int \frac{\frac{d}{d x} (\frac{d y}{d x})}{(\frac{d y}{d x})} = \int C_{4} d x \ln (\frac{d y}{d x}) = C_{4} x + C_{5} \Rightarrow \frac{d y}{d x} = e^{C_{4} x + C_{5}} \int d y = \int (e^{C_{4} x + C_{5}}) d x y = \frac{e^{C_{4} x + C_{5}}}{C_{4}} + C_{6} y = \frac{e^{C_{4} x} . e^{C_{5}}}{C_{4}} + C_{6} \Rightarrow y = C_{1} e^{C_{2} x} + C_{3} where, C_{1} = \frac{e^{C_{5}}}{C_{4}} C_{4} = C_{2} C_{6} = C_{3}$

Page No 22.140:

Question 15:

The general solution of the differential equation $\frac{d y}{d x} + y$ g' (x) = g (x) g' (x), where g (x) is a given function of x, is
(a) g (x) + log {1 + y + g (x)} = C
(b) g (x) + log {1 + y − g (x)} = C
(c) g (x) − log {1 + y − g (x)} = C
(d) none of these

Answer:

(b) g (x) + log {1 + y − g (x)} = C

$We have, \frac{d y}{d x} + y g' (x) = g (x) g' (x) . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = g' (x) and Q = g (x) g' (x) . ∴ I . F . = e^{\int P d x} = e^{\int g' (x) d x} = e^{g (x)} Multiplying both sides of (1) by I . F ., we get e^{g (x)} (\frac{d y}{d x} + y g' (x)) = e^{g (x)} g (x) g' (x) \Rightarrow e^{g (x)} \frac{d y}{d x} + e^{g (x)} y g' (x) = e^{g (x)} g (x) g' (x) Integrating both sides with respect to x, we get y e^{g (x)} = \int e^{g (x)} g (x) g' (x) d x + K \Rightarrow y e^{g (x)} = I + K where I = \int e^{g (x)} g (x) g' (x) d x Now, I = \int e^{g (x)} g (x) g' (x) d x Putting g (x) = t, we get g' (x) d x = d t ∴ I = \int \underset{I}{t} \underset{II}{e^{t}} d t = t \int e^{t} d t - \int [\frac{d}{d x} (t) \int e^{t} d t] d t = t e^{t} - e^{t} = g (x) e^{g (x)} - e^{g (x)} ∴ y e^{g (x)} = g (x) e^{g (x)} - e^{g (x)} + K \Rightarrow y e^{g (x)} + e^{g (x)} - g (x) e^{g (x)} = K \Rightarrow y + 1 - g (x) = K e^{- g (x)} Taking \log on both sides, we get \log \{y + 1 - g (x)\} = - g (x) + \log K \Rightarrow g (x) + \log \{1 + y - g (x)\} = C (Where, C = \log K)$

Page No 22.140:

Question 16:

The solution of the differential equation $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}, y (0) = 0$ is
(a) $y^{2} = \exp (x + \frac{x^{2}}{2}) - 1$

(b) $y^{2} = 1 + C \exp (x + \frac{x^{2}}{2})$

(c) y = tan (C + x + x²)

(d) $y = \tan (x + \frac{x^{2}}{2})$

Answer:

$(d) y = \tan (x + \frac{x^{2}}{2})$

We have,
$\frac{d y}{d x} = 1 + x + y^{2} + x y^{2} \Rightarrow \frac{d y}{d x} = (x + 1) + y^{2} (x + 1) \Rightarrow \frac{d y}{d x} = (x + 1) (1 + y^{2}) \Rightarrow \frac{d y}{(1 + y^{2})} = (x + 1) d x Integrating both sides, we get \int \frac{d y}{(1 + y^{2})} = \int (x + 1) d x \Rightarrow \tan^{- 1} y = \frac{x^{2}}{2} + x + C . . . . . (1) Now, y (0) = 0 ∴ \tan^{- 1} 0 = \frac{0}{2} + 0 + C \Rightarrow C = 0 Putting the value of C in (1), we get \tan^{- 1} y = \frac{x^{2}}{2} + x \Rightarrow y = \tan (\frac{x^{2}}{2} + x)$

Page No 22.141:

Question 17:

The differential equation of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = C$ is
(a) $\frac{y "}{y'} + \frac{y'}{y} - \frac{1}{x} = 0$

(b) $\frac{y "}{y'} + \frac{y'}{y} + \frac{1}{x} = 0$

(c) $\frac{y "}{y'} - \frac{y'}{y} - \frac{1}{x} = 0$

(d) none of these

Answer:

(a) $\frac{y "}{y'} + \frac{y'}{y} - \frac{1}{x} = 0$

We have,
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = C . . . . . (1)$
Differentiating with respect to x, we get
$\frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} y' = 0 \Rightarrow \frac{x}{a^{2}} + \frac{y}{b^{2}} y' = 0 . . . . . (2) Again differentiating with respect to x, we get \Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} {(y')}^{2} + \frac{y}{b^{2}} y'' = 0 . . . . . (3) Multiplying throughout by x, we get \frac{x}{a^{2}} + \frac{x}{b^{2}} {(y')}^{2} + \frac{x y}{b^{2}} y'' = 0 . . . . . (4) Subtracting (2) from (4), we get \frac{1}{b^{2}} [x {(y')}^{2} + x y y'' - y y'] = 0 \Rightarrow x {(y')}^{2} + x y y'' - y y' = 0 Dividing both sides by x y y', we get \frac{y'}{y} + \frac{y''}{y'} - \frac{1}{x} = 0 \Rightarrow \frac{y''}{y'} + \frac{y'}{y} - \frac{1}{x} = 0$

Page No 22.141:

Question 18:

Solution of the differential equation $\frac{d y}{d x} + \frac{y}{x}$ = sin x is
(a) x (y + cos x) = sin x + C
(b) x (y − cos x) = sin x + C
(c) x (y + cos x) = cos x + C
(d) none of these

Answer:

(a) x (y + cos x) = sin x + C

$We have, \frac{d y}{d x} + \frac{y}{x} = \sin x \Rightarrow \frac{d y}{d x} + \frac{1}{x} y = \sin x . . . . . (1) Comparing with \frac{d y}{d x} + P y = Q, we get P = \frac{1}{x} Q = \sin x Now, I . F . = e^{\int \frac{1}{x} d x} = e^{\log |x|} = x Therefore, integration of (1) is given by y \times I . F . = \int x^{2} \times I . F . d x + C \Rightarrow y x = \int \underset{I}{x} \underset{II}{\sin x} d x + C \Rightarrow y x = x \int \sin x d x - \int [\frac{d}{d x} (x) \int \sin x d x] d x + C \Rightarrow y x = - x \cos x + \int \cos x d x + C \Rightarrow y x + x \cos x = \sin x + C \Rightarrow x (y + \cos x) = \sin x + C$

Page No 22.141:

Question 19:

The equation of the curve satisfying the differential equation y (x + y³) dx = x (y³ − x) dy and passing through the point (1, 1) is
(a) y³ − 2x + 3x²y = 0
(b) y³ + 2x + 3x²y = 0
(c) y³ + 2x −3x²y = 0
(d) none of these

Answer:

(c) y³ + 2x −3x²y = 0

We have,

$y (x + y^{3}) d x = x (y^{3} - x) d y Here, (x y + y^{4}) d x = (x y^{3} - x^{2}) d y \Rightarrow x y d x + y^{4} d x - x y^{3} d y + x^{2} d y = 0 \Rightarrow x (y d x + x d y) + y^{3} (y d x - x d y) = 0 \Rightarrow x d (x y) + x^{2} y^{3} \frac{(y d x - x d y)}{x^{2}} = 0 \Rightarrow x d (x y) - x^{2} y^{3} \frac{(x d y - y d x)}{x^{2}} = 0 \Rightarrow \frac{d (x y)}{x^{2} y^{2}} - \frac{y}{x} d (\frac{y}{x}) = 0 (∵ Dividing the whole equation b y x^{3} y^{2}) \Rightarrow \frac{d (x y)}{x^{2} y^{2}} = \frac{y}{x} d (\frac{y}{x})$
Integrating both sides we get,
$\Rightarrow \int \frac{d (x y)}{x^{2} y^{2}} = \int \frac{y}{x} d (\frac{y}{x}) \Rightarrow - \frac{1}{x y} = \frac{{(\frac{y}{x})}^{2}}{2} - c ∴ - \frac{1}{x y} - \frac{1}{2} {(\frac{y}{x})}^{2} - c = 0 ∴ \frac{1}{x y} + \frac{1}{2} (\frac{y^{2}}{x^{2}}) + c = 0 ∴ y^{3} + 2 x + 2 c x^{2} y = 0$

It is given that the curves passes through (1, 1).
Hence,
$y^{3} + 2 x + 2 c x^{2} y = 0 {(1)}^{3} + 2 (1) + 2 c (1) (1) = 0 1 + 2 + 2 c = 0 2 c = - 3 c = - \frac{3}{2}$
∴ The required curve is $y^{3} + 2 x - 2 \times \frac{3}{2} x^{2} y = 0$
$∴ y^{3} + 2 x - 3 x^{2} y = 0$

Page No 22.141:

Question 20:

The solution of the differential equation $2 x \frac{d y}{d x} - y = 3$ represents
(a) circles
(b) straight lines
(c) ellipses
(d) parabolas

Answer:

(d) parabolas

We have,
$2 x \frac{d y}{d x} - y = 3 \Rightarrow 2 x \frac{d y}{d x} = 3 + y \Rightarrow \frac{1}{3 + y} d y = \frac{1}{2 x} d x Integrating both sides, we get \int \frac{1}{3 + y} d y = \frac{1}{2} \int \frac{1}{x} d x \Rightarrow \log |3 + y| = \frac{1}{2} \log |x| + \log C \Rightarrow \log |3 + y| - \log |x^{\frac{1}{2}}| = \log C \Rightarrow \log |\frac{3 + y}{\sqrt{x}}| = \log C \Rightarrow \frac{3 + y}{\sqrt{x}} = C \Rightarrow 3 + y = C \sqrt{x} Squaring both sides, we get {(3 + y)}^{2} = C x . . . . . (1) Thus, (1) represents the equation of parabolas .$

Page No 22.141:

Question 21:

The solution of the differential equation $x \frac{d y}{d x} = y + x \tan \frac{y}{x}$ , is
(a) $\sin \frac{x}{y} = x + C$

(b) $\sin \frac{y}{x} = C x$

(c) $\sin \frac{x}{y} = C y$

(d) $\sin \frac{y}{x} = C y$

Answer:

(b) $\sin \frac{y}{x} = C x$

We have,
$x \frac{d y}{d x} = y + x \tan \frac{y}{x} \Rightarrow \frac{d y}{d x} = \frac{y}{x} + \tan \frac{y}{x} . . . . . (1) Let y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} Putting the above value in (1), we get v + x \frac{d v}{d x} = v + \tan v \Rightarrow x \frac{d v}{d x} = \tan v \Rightarrow \frac{d v}{\tan v} = \frac{d x}{x} Integrating both sides, we get \log \sin v = \log x + \log C \Rightarrow \log \sin v - \log x = \log C \Rightarrow \log \frac{\sin v}{x} = \log C \Rightarrow \frac{\sin v}{x} = C \Rightarrow \sin v = C x \Rightarrow \sin (\frac{y}{x}) = C x [∵ y = v x]$

Page No 22.141:

Question 22:

The differential equation satisfied by ax² + by² = 1 is
(a) xyy₂+ y₁² + yy₁ = 0
(b) xyy₂ + xy₁² − yy₁ = 0
(c) xyy₂ − xy₁² + yy₁ = 0
(d) none of these

Answer:

(b) xyy₂ + xy₁² − yy₁ = 0

We have,
ax² + by² = 1 .....(1)
Differentiating both sides of (1) with respect to x, we get
$2 a x + 2 b y \frac{d y}{d x} = 0 . . . . . (2)$
Differentiating both sides of (2) with respect to x, we get
$2 a + 2 b {(\frac{d y}{d x})}^{2} + 2 b y \frac{d^{2} y}{d x^{2}} = 0 \Rightarrow 2 b [y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}] = - 2 a \Rightarrow [y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}] = - \frac{2 a}{2 b} \Rightarrow [y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}] = - (- \frac{y}{x} \frac{d y}{d x}) [Using (2)] \Rightarrow x [y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}] = y \frac{d y}{d x} \Rightarrow x y \frac{d^{2} y}{d x^{2}} + x {(\frac{d y}{d x})}^{2} = y \frac{d y}{d x} \Rightarrow x y \frac{d^{2} y}{d x^{2}} + x {(\frac{d y}{d x})}^{2} - y \frac{d y}{d x} = 0 \Rightarrow x y y_{2} + x {(y_{1})}^{2} - y y_{1} = 0$

Page No 22.141:

Question 23:

The differential equation which represents the family of curves y = e^Cx is
(a) y₁ = C²y
(b) xy₁ − ln y = 0
(c) x ln y = yy₁
(d) y ln y = xy₁

Answer:

(d) y ln y = xy₁

We have,
y = e^Cx
Taking ln on both sides, we get
ln y = Cx ln e
$\Rightarrow \ln y = C x . . . . . (1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{1}{y} y_{1} = C$
Substituting the value of C in (1), we get
$\ln y = \frac{y_{1}}{y} x \Rightarrow y \ln y = y_{1} x$

Page No 22.141:

Question 24:

Which of the following transformations reduce the differential equation $\frac{d z}{d x} + \frac{z}{x} \log z = \frac{z}{x^{2}} {(\log z)}^{2}$ into the form $\frac{d u}{d x} + P (x) u = Q (x)$
(a) u = log x
(b) u = e^z
(c) u = (log z)⁻¹
(d) u = (log z)²

Answer:

(c) u = (log z)⁻¹

$Given \frac{d z}{d x} + \frac{z}{x} \log z = \frac{z}{x^{2}} {(\log z)}^{2} . . . . . (1) Let u = {(\log z)}^{- 1} \frac{d u}{d x} = - \frac{1}{{(\log z)}^{2}} \times \frac{1}{z} \times \frac{d z}{d x} \frac{d z}{d x} = - z {(\log z)}^{2} \frac{d u}{d x} Substituting the value of \frac{d z}{d x} from equation (1) we get, ∴ - z {(\log z)}^{2} \frac{d u}{d x} + \frac{z}{x} \log z = \frac{z}{x^{2}} {(\log z)}^{2} \frac{d u}{d x} - \frac{1}{x} \frac{1}{\log z} = - \frac{1}{x^{2}} \frac{d u}{d x} - \frac{1}{x} {(\log z)}^{- 1} = - \frac{1}{x^{2}} \frac{d u}{d x} - \frac{1}{x} u = - \frac{1}{x^{2}} It can be written as, \frac{d u}{d x} + p (x) u = Q (x) w here, p (x) = - \frac{1}{x} q (x) = - \frac{1}{x^{2}}$

The correct option is C.

Page No 22.141:

Question 25:

The solution of the differential equation $\frac{d y}{d x} = \frac{y}{x} + \frac{ϕ (\frac{y}{x})}{ϕ' (\frac{y}{x})}$ is
(a) $ϕ (\frac{y}{x}) = k x$

(b) $x ϕ (\frac{y}{x}) = k$

(c) $ϕ (\frac{y}{x}) = k y$

(d) $y ϕ (\frac{y}{x}) = k$

Answer:

$(a) ϕ (\frac{y}{x}) = k x$

$We have, \frac{d y}{d x} = \frac{y}{x} + \frac{ϕ (\frac{y}{x})}{ϕ' (\frac{y}{x})} Let y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} ∴ v + x \frac{d v}{d x} = v + \frac{ϕ (v)}{ϕ' (v)} \Rightarrow x \frac{d v}{d x} = \frac{ϕ (v)}{ϕ' (v)} \Rightarrow \frac{ϕ (v)}{ϕ' (v)} d v = \frac{1}{x} d x Integrating both sides, we get \int \frac{ϕ' (v)}{ϕ (v)} d v = \int \frac{1}{x} d x \Rightarrow \log |ϕ (v)| = \log |x| + \log k \Rightarrow \log |ϕ (\frac{y}{x})| - \log |x| = \log k \Rightarrow \log |\frac{ϕ (\frac{y}{x})}{x}| = \log k \Rightarrow \frac{ϕ (\frac{y}{x})}{x} = k \Rightarrow ϕ (\frac{y}{x}) = kx$

Page No 22.141:

Question 26:

If m and n are the order and degree of the differential equation ${(y_{2})}^{5} + \frac{4 {(y_{2})}^{3}}{y_{3}} + y_{3} = x^{2} - 1$ , then
(a) m = 3, n = 3
(b) m = 3, n = 2
(c) m = 3, n = 5
(d) m = 3, n = 1

Answer:

(b) m = 3, n = 2

$We have, {(y_{2})}^{5} + \frac{4 {(y_{2})}^{3}}{y_{3}} + y_{3} = x^{2} - 1 \Rightarrow y_{3} {(y_{2})}^{5} + 4 {(y_{2})}^{3} + {(y_{3})}^{2} = y_{3} (x^{2} - 1) The highest order derivative is y_{3} and its highest exponent in this equation is 2 . Therefore, order is 3 and degree is 2 . Hence, m = 3, n = 2$

Page No 22.142:

Question 27:

The solution of the differential equation $\frac{d y}{d x} + 1 = e^{x + y}$ , is
(a) (x + y) e^x^{+ y} = 0
(b) (x + C) e^x^{+ y} = 0
(c) (x − C) e^x^{+ y} = 1
(d) (x − C) e^x^{+ y}+ 1 =0

Answer:

(d) (x − C) e^x^{+ y}+ 1 = 0

$We have, \frac{d y}{d x} + 1 = e^{x + y} Let x + y = v \Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \Rightarrow \frac{d y}{d x} + 1 = \frac{d v}{d x} ∴ \frac{d v}{d x} = e^{v} \Rightarrow e^{- v} d v = d x Integrating both sides, we get - e^{- v} = x - C \Rightarrow - 1 = e^{v} (x - C) \Rightarrow (x - C) e^{x + y} + 1 = 0$

Page No 22.142:

Question 28:

The solution of x² + y² $\frac{d y}{d x}$ = 4, is
(a) x² + y² = 12x + C
(b) x² + y² = 3x + C
(c) x³ + y³ = 3x + C
(d) x³ + y³ = 12x + C

Answer:

(d) x³ + y³ = 12x + C

$We have, x^{2} + y^{2} \frac{d y}{d x} = 4 \Rightarrow y^{2} \frac{d y}{d x} = 4 - x^{2} \Rightarrow y^{2} d y = (4 - x^{2}) d x Integrating both sides, we get \int y^{2} d y = \int (4 - x^{2}) d x \Rightarrow \frac{y^{3}}{3} = 4 x - \frac{x^{3}}{3} + D \Rightarrow y^{3} = 12 x - x^{3} + 3 D \Rightarrow x^{3} + y^{3} = 12 x + C, where C = 3 D$

Page No 22.142:

Question 29:

The family of curves in which the sub tangent at any point of a curve is double the abscissae, is given by
(a) x = Cy²
(b) y = Cx²
(c) x² = Cy²
(d) y = Cx

Answer:

(a) x = Cy²

$Subtangent = \frac{y}{\frac{d y}{d x}}$

It is given that subtangent at any point of a curve is double of the abscissa.

$\begin{array}{l} ∴ \frac{y}{\frac{d y}{d x}} = 2 x \\ y = 2 x \frac{d y}{d x} \\ \int \frac{d x}{x} = 2 \int \frac{d y}{y} \\ \ln x = 2 \ln y + a \\ \ln x = \ln y^{2} + \ln c \\ \ln x = \ln c y^{2} \\ x = c y^{2} \end{array}$

Page No 22.142:

Question 30:

The solution of the differential equation x dx + y dy = x²y dy − y²x dx, is
(a) x² − 1 = C (1 + y²)
(b) x² + 1 = C (1 − y²)
(c) x³ − 1 = C (1 + y³)
(d) x³ + 1 = C (1 − y³)

Answer:

(a) x² − 1 = C (1 + y²)

We have,
x dx + y dy = x²y dy − y²x dx

$\Rightarrow (x + x y^{2}) d x = (x^{2} y - y) d y \Rightarrow \frac{x}{(x^{2} - 1)} d x = \frac{y}{(1 + y^{2})} d y \Rightarrow \frac{2 x}{2 (x^{2} - 1)} d x = \frac{2 y}{2 (1 + y^{2})} d y Integrating both sides, we get \frac{1}{2} \int \frac{2 y}{(1 + y^{2})} d y = \frac{1}{2} \int \frac{2 x}{(x^{2} - 1)} d x \Rightarrow \frac{1}{2} \log |(1 + y^{2})| = \frac{1}{2} \log |(x^{2} - 1)| - \frac{1}{2} \log |C| \Rightarrow \log |(1 + y^{2})| = \log |(x^{2} - 1)| - \log |C| \Rightarrow \log |(1 + y^{2})| = \log |(\frac{x^{2} - 1}{C})| \Rightarrow 1 + y^{2} = \frac{x^{2} - 1}{C} \Rightarrow C (1 + y^{2}) = x^{2} - 1$

Page No 22.142:

Question 31:

The solution of the differential equation (x² + 1) $\frac{d y}{d x}$ + (y² + 1) = 0, is
(a) y = 2 + x²

(b) $y = \frac{1 + x}{1 - x}$

(c) y = x (x − 1)

(d) $y = \frac{1 - x}{1 + x}$

Answer:

$(d) y = \frac{1 - x}{1 + x}$

$We have, (x^{2} + 1) \frac{d y}{d x} + (y^{2} + 1) = 0 \Rightarrow (x^{2} + 1) \frac{d y}{d x} = - (y^{2} + 1) \Rightarrow \frac{1}{(y^{2} + 1)} d y = - \frac{1}{(x^{2} + 1)} d x Integrating both sides, we get \int \frac{1}{(y^{2} + 1)} d y = - \int \frac{1}{(x^{2} + 1)} d x \Rightarrow \tan^{- 1} y = - \tan^{- 1} x + \tan^{- 1} C \Rightarrow \tan^{- 1} y + \tan^{- 1} x = \tan^{- 1} C \Rightarrow \tan^{- 1} (\frac{x + y}{1 - x y}) = \tan^{- 1} C \Rightarrow \frac{x + y}{1 - x y} = C D i s c l a i m e r : T h e i n i t i a l v a l u e c o n d i t i o n s a r e n o t g i v e n, s o t h e f i n a l a n s w e r w i l l b e o b a t i n e d o n l y i f C = 1 . S o, \Rightarrow x + y = 1 - x y \Rightarrow y + x y = 1 - x \Rightarrow y (1 + x) = 1 - x \Rightarrow y = \frac{1 - x}{1 + x}$

Page No 22.142:

Question 32:

The differential equation $x \frac{d y}{d x} - y = x^{2}$ , has the general solution
(a) y − x³ = 2cx
(b) 2y − x³ = cx
(c) 2y + x² = 2cx
(d) y + x² = 2cx

Answer:

(b) 2y − x³ = cx

We have,

$x \frac{d y}{d x} - y = x^{2}$

$\Rightarrow \frac{d y}{d x} - \frac{1}{x} y = x^{2} Comparing with \frac{d y}{d x} + P y = Q, we get P = - \frac{1}{x} Q = x^{2} Now, I . F . = e^{- \int \frac{1}{x} d x} = e^{- \log |x|} = e^{\log |\frac{1}{x}|} = \frac{1}{x} y \times I . F = \int x^{2} \times I . Fd x + C \Rightarrow y \frac{1}{x} = \int x^{2} \times \frac{1}{x} d x + C \Rightarrow y \frac{1}{x} = \int x d x + C \Rightarrow y \frac{1}{x} = \frac{x^{2}}{2} + C \Rightarrow 2 y - x^{3} = C x$

Page No 22.142:

Question 33:

The solution of the differential equation $\frac{d y}{d x} - k y = 0, y (0) = 1$ approaches to zero when x → ∞, if
(a) k = 0
(b) k > 0
(c) k < 0
(d) none of these

Answer:

(c) k < 0

$We have, \Rightarrow \frac{d y}{d x} - k y = 0 \Rightarrow \frac{d y}{d x} = k y \Rightarrow \frac{1}{y} d y = k d x Integrating both sides, we get \int \frac{1}{y} d y = k \int d x \Rightarrow \log |y| = k x + C . . . . . (1) Now, y (0) = 1 ∴ C = 0 Putting C = 0 in (1), we get \log |y| = k x \Rightarrow e^{k x} = y According to the question, e^{k \infty} = 0 Since e^{- \infty} = 0 ∴ k < 0 .$

Page No 22.142:

Question 34:

The solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + 1 + y^{2} = 0$ , is
(a) tan⁻¹x − tan⁻¹y = tan⁻¹C
(b) tan⁻¹y − tan⁻¹x = tan⁻¹C
(c) tan⁻¹y ± tan⁻¹x = tan C
(d) tan⁻¹y + tan⁻¹x = tan⁻¹C

Answer:

(d) tan⁻¹y + tan⁻¹x = tan⁻¹C

We have,
$(1 + x^{2}) \frac{d y}{d x} + 1 + y^{2} = 0 \Rightarrow (1 + x^{2}) \frac{d y}{d x} = - (1 + y^{2}) \Rightarrow \frac{1}{(1 + y^{2})} d y = - \frac{1}{(1 + x^{2})} d x Integrating both sides we get, \int \frac{1}{(1 + y^{2})} d y = - \int \frac{1}{(1 + x^{2})} d x \Rightarrow \tan^{- 1} y = - \tan^{- 1} x + \tan^{- 1} C \Rightarrow \tan^{- 1} y + \tan^{- 1} x = \tan^{- 1} C$

Page No 22.142:

Question 35:

The solution of the differential equation $\frac{d y}{d x} = \frac{x^{2} + x y + y^{2}}{x^{2}}$ , is
(a) $\tan^{- 1} (\frac{x}{y}) = \log y + C$

(b) $\tan^{- 1} (\frac{y}{x}) = \log x + C$

(c) $\tan^{- 1} (\frac{x}{y}) = \log x + C$

(d) $\tan^{- 1} (\frac{y}{x}) = \log y + C$

Answer:

$(b) \tan^{- 1} (\frac{y}{x}) = \log x + C$

We have,
$\frac{d y}{d x} = \frac{x^{2} + x y + y^{2}}{x^{2}} (1)$
This is homogenous differential equation.
$Let y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} Now, putting \frac{d y}{d x} = v + x \frac{d v}{d x} and y = v x in (1), we get v + x \frac{d v}{d x} = \frac{x^{2} + x^{2} v + x^{2} v^{2}}{x^{2}} \Rightarrow v + x \frac{d v}{d x} = 1 + v + v^{2} \Rightarrow x \frac{d v}{d x} = 1 + v^{2} \Rightarrow (\frac{1}{1 + v^{2}}) d v = \frac{1}{x} d x Integrating both sides we get, \int \frac{1}{1 + v^{2}} d v = \int \frac{1}{x} d x \Rightarrow \tan^{- 1} v = \log x + C \Rightarrow \tan^{- 1} (\frac{y}{x}) = \log x + C$

Page No 22.142:

Question 36:

The differential equation $\frac{d y}{d x} + P y = Q y^{n}, n > 2$ can be reduced to linear form by substituting
(a) z = yⁿ⁻¹
(b) z = yⁿ
(c) z = yⁿ^{+ 1}
(d) z = y¹^{− n}

Answer:

(d) z = y¹^{− n}

We have,
$\frac{d y}{d x} + P y = Q y^{n} \Rightarrow y^{- n} \frac{d y}{d x} + P y^{1 - n} = Q . . . . . (1) Put z = y^{1 - n} Integrating both sides with respect to x, we get \frac{d z}{d x} = (1 - n) y^{- n} \frac{d y}{d x} \Rightarrow y^{- n} \frac{d y}{d x} = \frac{1}{(1 - n)} \frac{d z}{d x} Now, (1) becomes \frac{1}{(1 - n)} \frac{d z}{d x} + P z = Q \Rightarrow \frac{d z}{d x} + P (1 - n) z = Q (1 - n) Which is linear form of differential equation . Therefore, the given differential equation can be reduce to linear form by the substitution, z = y^{1 - n}$

Page No 22.142:

Question 37:

If p and q are the order and degree of the differential equation $y \frac{d y}{d x} + x^{3} \frac{d^{2} y}{d x^{2}} + x y$ = cos x, then
(a) p < q
(b) p = q
(c) p > q
(d) none of these

Answer:

(c) p > q

We have,
$y \frac{d y}{d x} + x^{3} \frac{d^{2} y}{d x^{2}} + x y = \cos x$
$The highest order derivative is \frac{d^{2} y}{d^{2} x} and it' s degree is 1 So, the order is 2 and the degree is 1 . ∴ p = 2 and q = 1 Clearly, p > q$

Page No 22.143:

Question 38:

Which of the following is the integrating factor of (x log x) $\frac{d y}{d x} + y$ = 2 log x?
(a) x
(b) e^x
(c) log x
(d) log (log x)

Answer:

(c) log x

We have,
$(x \log x) \frac{d y}{d x} + y = 2 \log x$
Dividing both sides by (x log x) we get,
$\frac{d y}{d x} + \frac{y}{x \log x} = 2 \frac{\log x}{x \log x} \Rightarrow \frac{d y}{d x} + \frac{y}{x \log x} = \frac{2}{x} \Rightarrow \frac{d y}{d x} + (\frac{1}{x \log x}) y = \frac{2}{x} Comparing with \frac{d y}{d x} + P y = Q we get, P = \frac{1}{x \log x} and Q = \frac{2}{x} Now, I . F = e^{\int P d x} = e^{\int \frac{1}{x \log x} d x} = e^{\log (\log x)} = \log x$

Page No 22.143:

Question 39:

What is integrating factor of $\frac{d y}{d x}$ + y sec x = tan x?
(a) sec x + tan x
(b) log (sec x + tan x)
(c) e^sec^x
(d) sec x

Answer:

(a) sec x + tan x

We have,

$\frac{d y}{d x} + y \sec x = \tan x Comparing with \frac{d y}{d x} + P y = Q, we get P = \sec x Q = \tan x Now, I . F . = e^{\int \sec x d x} = e^{\log (\sec x + \tan x)} = \sec x + \tan x$

Page No 22.143:

Question 40:

Integrating factor of the differential equation cos $x \frac{d y}{d x} + y \sin x = 1$ , is
(a) cos x
(b) tan x
(c) sec x
(d) sin x

Answer:

(c) sec x

We have,
$\cos x \frac{d y}{d x} + y \sin x = 1$
Dividing both sides by cos x, we get
$\frac{d y}{d x} + \frac{\sin x}{\cos x} y = \frac{1}{\cos x} \Rightarrow \frac{d y}{d x} + (\tan x) y = \frac{1}{\cos x} Comparing with \frac{d y}{d x} + P y = Q, we get P = \tan x Q = \frac{2}{\cos x} Now, I . F . = e^{\int \tan x d x} = e^{\log (s e c x)} = \sec x$

Page No 22.143:

Question 41:

The degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d y}{d x})}^{2} + \sin (\frac{d y}{d x}) + 1 = 0$ , is
(a) 3
(b) 2
(c) 1
(d) not defined

Answer:

(d) not defined

We have,
${(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d y}{d x})}^{2} + \sin (\frac{d y}{d x}) + 1 = 0$
$The highest order derivative in this equation is \frac{d^{2} y}{d^{2} x} . But the equation cannot be expressed as a polynomial in differential coefficient . Hence, the degree is not defined .$

Page No 22.143:

Question 42:

The order of the differential equation $2 x^{2} \frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + y = 0$ , is
(a) 2
(b) 1
(c) 0
(d) not defined

Answer:

(a) 2

We have,
$2 x^{2} \frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + y = 0$
$Here, the highest order derivative is \frac{d^{2} y}{d^{2} x} . Hence, the order is 2 .$

Page No 22.143:

Question 43:

The number of arbitrary constants in the general solution of differential equation of fourth order is
(a) 0
(b) 2
(c) 3
(d) 4

Answer:

(d) 4

The number of arbitrary constants in the general solution of a differential equation of order n is n.
Thus, the number of arbitrary constants in the general solution of differential equation of fourth order is 4.

Page No 22.143:

Question 44:

The number of arbitrary constants in the particular solution of a differential equation of third order is
(a) 3
(b) 2
(c) 1
(d) 0

Answer:

(d) 0

The number of arbitrary constants in the particular solution of a differential equation is always zero.

Page No 22.143:

Question 45:

Which of the following differential equations has y = C₁e^x + C₂e⁻^x as the general solution?
(a) $\frac{d^{2} y}{d x^{2}} + y = 0$

(b) $\frac{d^{2} y}{d x^{2}} - y = 0$

(c) $\frac{d^{2} y}{d x^{2}} + 1 = 0$

(d) $\frac{d^{2} y}{d x^{2}} - 1 = 0$

Answer:

$(b) \frac{d^{2} y}{d x^{2}} - y = 0$

We have,
$y = C_{1} e^{x} + C_{2} e^{- x} . . . . . (1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{d y}{d x} = C_{1} e^{x} - C_{2} e^{- x} . . . . . (2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{d^{2} y}{d x^{2}} = C_{1} e^{x} + C_{2} e^{- x} \Rightarrow \frac{d^{2} y}{d x^{2}} = y [Using (1) and (2)] \Rightarrow \frac{d^{2} y}{d x^{2}} - y = 0$

Page No 22.143:

Question 46:

Which of the following differential equations has y = x as one of its particular solution?
(a) $\frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = x$

(b) $\frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = x$

(c) $\frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = 0$

(d) $\frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = 0$

Answer:

$(c) \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = 0$

We have,
y = x .....(1)
Differentiating both sides of (1) with respect to x, we get
$\frac{d y}{d x} = 1 . . . . . (2) Differentiating again with respect to x, we get \Rightarrow \frac{d^{2} y}{d x^{2}} = 0 \Rightarrow \frac{d^{2} y}{d x^{2}} + x^{2} = x^{2} \Rightarrow \frac{d^{2} y}{d x^{2}} + x \times x = x^{2} \times 1 \Rightarrow \frac{d^{2} y}{d x^{2}} + x y = x^{2} \times 1 [Using (1)] \Rightarrow \frac{d^{2} y}{d x^{2}} + x y = x^{2} \frac{d y}{d x} [Using (2)] \Rightarrow \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = 0$

Page No 22.143:

Question 47:

The general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ , is
(a) e^x+ e^−y = C
(b) e^x + e^y = C
(c) e⁻^x + e^y = C
(d) e^−x + e^−y = C

Answer:

(a) e^x+ e^−y = C

We have,
$\frac{d y}{d x} = e^{x + y} \Rightarrow \frac{d y}{d x} = e^{x} \times e^{y} \Rightarrow e^{- y} d y = e^{x} d x Integrating both sides, we get \int e^{- y} d y = \int e^{x} d x \Rightarrow - e^{- y} = e^{x} + D \Rightarrow e^{x} + e^{- y} = - D \Rightarrow e^{x} + e^{- y} = C [Where, C = - D]$

Page No 22.143:

Question 48:

A homogeneous differential equation of the form $\frac{d x}{d y} = h (\frac{x}{y})$ can be solved by making the substitution
(a) y = vx
(b) v = yx
(c) x = vy
(d) x = v

Answer:

(c) x = vy

A homogeneous differential equation of the form $\frac{d x}{d y} = h (\frac{x}{y})$ can be solved by substituting x = vy.

Page No 22.143:

Question 49:

Which of the following is a homogeneous differential equation?
(a) (4x + 6y + 5) dy − (3y + 2x + 4) dx = 0
(b) xy dx − (x³ + y³) dy = 0
(c) (x³ + 2y²) dx + 2xy dy = 0
(d) y²dx + (x² − xy − y²) dy = 0

Answer:

(d) y²dx + (x² − xy − y²) dy = 0

A differential equation is said to be homogenous if all the terms in the equation have equal degree and it can be written in the form $\frac{d y}{d x} = \frac{f (x, y)}{g (x, y)}$ .

In (a), (b) and (c), the degree of all the terms is not equal.
But in the equation y²dx + (x² − xy − y²) dy = 0, the degree of all the terms is 2.
Thus, (d) contains a homogeneous differential equation.

Page No 22.143:

Question 50:

The integrating factor of the differential equation $x \frac{d y}{d x} - y = 2 x^{2}$
(a) e^−x

(b) e⁻^y

(c) $\frac{1}{x}$

(d) x

Answer:

$(c) \frac{1}{x}$

We have,

$x \frac{d y}{d x} - y = 2 x^{2} \Rightarrow \frac{d y}{d x} - \frac{1}{x} y = 2 x Comparing with \frac{d y}{d x} + P y = Q, we get P = - \frac{1}{x} Q = 2 x Now, I . F . = e^{- \int \frac{1}{x} d y} = e^{- \log |x|} = e^{\log |\frac{1}{x}|} = \frac{1}{x}$

Page No 22.144:

Question 51:

The integrating factor of the differential equation $(1 - y^{2}) \frac{d x}{d y} + y x = a y (- 1 < y < 1)$ is
(a) $\frac{1}{y^{2} - 1}$

(b) $\frac{1}{\sqrt{y^{2} - 1}}$

(c) $\frac{1}{1 - y^{2}}$

(d) $\frac{1}{\sqrt{1 - y^{2}}}$

Answer:

$(d) \frac{1}{\sqrt{1 - y^{2}}}$

$We have, (1 - y^{2}) \frac{d x}{d y} + y x = a y \frac{d x}{d y} + \frac{y}{1 - y^{2}} x = \frac{a y}{1 - y^{2}} Comparing with \frac{d x}{d y} + P x = Q, we get P = \frac{y}{1 - y^{2}} Q = \frac{a y}{1 - y^{2}} Now, I . F . = e^{\int \frac{y}{1 - y^{2}} d y} = e^{- \frac{1}{2} \int \frac{- 2 y}{1 - y^{2}} d y} = e^{- \frac{1}{2} \log |1 - y^{2}|} = e^{\log |\frac{1}{{\sqrt{1 - y}}^{2}}|} = \frac{1}{{\sqrt{1 - y}}^{2}}$

Page No 22.144:

Question 52:

The general solution of the differential equation $\frac{y d x - x d y}{y} = 0$ , is
(a) xy = C
(b) x = Cy²
(c) y = Cx
(d) y = Cx²

Answer:

(c) y = Cx

$We have, \frac{y d x - x d y}{y} = 0 \Rightarrow y d x = x d y \Rightarrow \frac{1}{y} d y = \frac{1}{x} d x Integrating both sides, we get \int \frac{1}{y} d y = \int \frac{1}{x} d x \Rightarrow \log y = \log x + D \Rightarrow \log y - \log x = \log C \Rightarrow \log (\frac{y}{x}) = \log C \Rightarrow \frac{y}{x} = C \Rightarrow y = C x$

Page No 22.144:

Question 53:

The general solution of a differential equation of the type $\frac{d x}{d y} + P_{1} x = Q_{1}$ is
(a) $y e^{\int P_{1} d y} = \int \{Q_{1} e^{\int P_{1} d y}\} d y + C$

(b) $y e^{\int P_{1} d x} = \int \{Q_{1} e^{\int P_{1} d x}\} d x + C$

(c) $x e^{\int P_{1} d y} = \int \{Q_{1} e^{\int P_{1} d y}\} d y + C$

(d) $x e^{\int P_{1} d x} = \int \{Q_{1} e^{\int P_{1} d x}\} d x + C$

Answer:

$(c) x e^{\int P_{1} d y} = \int \{Q_{1} e^{\int P_{1} d y}\} d y + C$

We have,
$\frac{d x}{d y} + P_{1} x = Q_{1}$
Comparing with the equation $\frac{d x}{d y} + P x = Q$ , we get
P = P₁
Q = Q₁
The general solution of the equation $\frac{d x}{d y} + P x = Q$ is given by
$x e^{\int P d y} = \int \{Q e^{\int P d y}\} d y + C$ ...(1)
Putting the value of P and Q in (1), we get
$x e^{\int P_{1} d y} = \int \{Q_{1} e^{\int P_{1} d y}\} d y + C$

Page No 22.144:

Question 54:

The general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0 is
(a) x e^y + x² = C
(b) x e^y + y² = C
(c) y e^x + x² = C
(d) y e^y + x² = C

Answer:

(c) y e^x + x² = C

We have,
e^x dy + (ye^x + 2x) dx = 0
$Dividing both sides by e^{x} d x, we get \frac{d y}{d x} + (y + \frac{2 x}{e^{x}}) = 0 \Rightarrow \frac{d y}{d x} + y = - \frac{2 x}{e^{x}} Comparing with \frac{d y}{d x} + P y = Q, we get P = 1 Q = - \frac{2 x}{e^{x}} Now, I . F . = e^{\int d x} = e^{x} Solution is given by, y \times I . F . = \int (Q \times I . F .) d x + C \Rightarrow y e^{x} = - \int e^{x} \times \frac{2 x}{e^{x}} d x + C \Rightarrow y e^{x} = - 2 \int x d x + C \Rightarrow y e^{x} = - x^{2} + C \Rightarrow y e^{x} + x^{2} = C$

Page No 22.144:

Question 1:

Determine the order and degree (if defined) of the following differential equations:
(i) ${(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0$

(ii) y"' + 2y" + y' = 0
(iii) (y"')² + (y")³ + (y')⁴ + y⁵ = 0
(iv) y"' + 2y" + y' = 0
(v) y" + (y')² + 2y = 0
(vi) y" + 2y' + sin y = 0
(vii) y"' + y² + e^y^' = 0

Answer:

$(i) {(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0$
The highest order derivative in the given equation is $\frac{d^{2} s}{d t^{2}}$ and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(ii) y"' + 2y" + y' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order and first degree.
i.e., Order = 3 and degree = 1

(iii) (y"')² + (y")³ + (y')⁴ + y⁵ = 0
The highest order derivative in the given equation is y''' and its power is 2.
Therefore, the given differential equation is of third order and second degree.
i.e., Order = 3 and degree = 2

(iv) y"' + 2y" + y' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order and first degree.
i.e., Order = 3 and degree = 1

(v) y" + (y')² + 2y = 0
The highest order derivative in the given equation is y'' and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(vi) y" + 2y' + sin y = 0
The highest order derivative in the given equation is y'' and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(vii) y"' + y² + e^y^' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order. This equation cannot be expressed as a polynomial of derivative.
Thus, the degree is not defined.
i.e., Order = 3 and degree is not defined.

Page No 22.144:

Question 2:

Verify that the function y = e^−3x is a solution of the differential equation $\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} - 6 y = 0$ .

Answer:

^{$We have, \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} - 6 y = 0 . . . . . (1) Now, y = e^{- 3 x} \Rightarrow \frac{d y}{d x} = - 3 e^{- 3 x} \Rightarrow \frac{d^{2} y}{d x^{2}} = 9 e^{- 3 x}$}
$Putting the values of \frac{d^{2} y}{d x^{2}}, \frac{d y}{d x} and y in (1), we get$

$LHS = 9 e^{- 3 x} - 3 e^{- 3 x} - 6 e^{- 3 x} = 0 = RHS$

Thus, y = e^−3x is the solution of the given differential equation.

Page No 22.144:

Question 3:

In each of the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

(i) y = e^x + 1	y'' − y' = 0
(ii) y = x² + 2x + C	y' − 2x − 2 = 0
(iii) y = cos x + C	y' + sin x = 0
(iv) y = $\sqrt{1 + x^{2}}$	y' = $\frac{x y}{1 + x^{2}}$
(v) y = x sin x	xy' = y + x $\sqrt{x^{2} - y^{2}}$
(vi) $y = \sqrt{a^{2} - x^{2}}$	$x + y \frac{d y}{d x} = 0$

Answer:

(i) We have,
y'' − y' = 0                            .....(1)
Now,
y = e^x +1

^{$\Rightarrow y' = e^{x} \Rightarrow y'' = e^{x}$}

Putting the above values in (1), we get
$LHS = e^{x} - e^{x} = 0 = RHS$
Thus, y = e^x +1 is the solution of the given differential equation.

(ii) We have,
y' − 2x − 2 = 0                    .....(1)
Now,
y = x² + 2x + C
^{$\Rightarrow y' = 2 x + 2$}
Putting the above value in (1), we get
$LHS = 2 x + 2 - 2 x - 2 = 0 = RHS$
Thus, y = x² + 2x + C is the solution of the given differential equation.

(iii) We have,
y' + sin x = 0    .....(1)
Now,
y = cos x + C
^{$\Rightarrow y' = - \sin x$}
Putting the above value in (1), we get
$LHS = - \sin x + \sin x = 0 = RHS$
Thus, y = cos x + C is the solution of the given differential equation.

(iv) We have,
y' = $\frac{x y}{1 + x^{2}}$                   .....(1)
Now,
y = $\sqrt{1 + x^{2}}$
^{$\Rightarrow y' = \frac{x}{\sqrt{1 + x^{2}}}$}

Putting the above value in (1), we get
$LHS = \frac{x}{\sqrt{1 + x^{2}}} = \frac{x}{\sqrt{1 + x^{2}}} \times \frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}} = \frac{x y}{1 + x^{2}} = RHS$
Thus, y = $\sqrt{1 + x^{2}}$ is the solution of the given differential equation.

(v) We have,
xy' = y + x $\sqrt{x^{2} - y^{2}}$      .....(1)
Now,
y = x sin x
^{$\Rightarrow y' = \sin x + x \cos x$}
Putting the above value in (1), we get
$LHS = x (\sin x + x \cos x) = x \sin x + x^{2} \cos x = x \sin x + x (x \cos x) = x \sin x + x (x \sqrt{1 - \sin^{2} x}) = x \sin x + x (\sqrt{x^{2} - x^{2} \sin^{2} x}) = y + x (\sqrt{x^{2} - y^{2}}) = RHS$
Thus, y = x sin x is the solution of the given differential equation.

(v) We have,
xy' = y + x $\sqrt{x^{2} - y^{2}}$ .....(1)
Now,
y = x sin x
^{$\Rightarrow y' = \sin x + x \cos x$}
Putting the above value in (1), we get
$LHS = x (\sin x + x \cos x) = x \sin x + x^{2} \cos x = x \sin x + x (x \cos x) = x \sin x + x (x \sqrt{1 - \sin^{2} x}) = x \sin x + x (\sqrt{x^{2} - x^{2} \sin^{2} x}) = y + x (\sqrt{x^{2} - y^{2}}) = RHS$
Thus, y = x sin x is the solution of the given differential equation.

(vi) We have,
$x + y \frac{d y}{d x} = 0$    .....(1)

Now,
$y = \sqrt{a^{2} - x^{2}}$
^{$\Rightarrow y' = \frac{- x}{\sqrt{a^{2} - x^{2}}}$}

Putting the above value in (1), we get

$LHS = x + y (\frac{- x}{\sqrt{a^{2} - x^{2}}}) = x + \sqrt{a^{2} - x^{2}} (\frac{- x}{\sqrt{a^{2} - x^{2}}}) = x + \sqrt{a^{2} - x^{2}} (\frac{- x}{\sqrt{a^{2} - x^{2}}}) = x - x = 0 = RHS$

Thus, $y = \sqrt{a^{2} - x^{2}}$ is the solution of the given differential equation.

Page No 22.145:

Question 4:

Form the differential equation representing the family of curves y = mx, where m is an arbitrary constant.

Answer:

We have,
y = mx (1)
Differentiating both sides, we get
$\frac{d y}{d x} = m \Rightarrow \frac{d y}{d x} = \frac{y}{x} [From (1)] \Rightarrow x \frac{d y}{d x} = y \Rightarrow x \frac{d y}{d x} - y = 0$

Page No 22.145:

Question 5:

Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constant.

Answer:

We have,
y = a sin (x + b) .....(2)
Differentiating both sides, we get
$\frac{d y}{d x} = a \cos (x + b) \Rightarrow \frac{d^{2} y}{d x^{2}} = - a \sin (x + b) \Rightarrow \frac{d^{2} y}{d x^{2}} = - a \times \frac{y}{a} [Using (2)] \Rightarrow \frac{d^{2} y}{d x^{2}} = - y \Rightarrow \frac{d^{2} y}{d x^{2}} + y = 0$

Page No 22.145:

Question 6:

Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.

Answer:

The equation of the parabola having vertex at origin and axis along the positive direction of x-axis is given by
y²=4ax .....(1)
Since there is only one parameter, so we differentiate it only once.
Differentiating with respect to x, we get
$2 y \frac{d y}{d x} = 4 a$
Substituting the value of 4a in (1), we get
$y^{2} = 2 y \frac{d y}{d x} \times x \Rightarrow y^{2} = 2 x y \frac{d y}{d x} \Rightarrow y^{2} - 2 x y \frac{d y}{d x} = 0$

Page No 22.145:

Question 7:

Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.

Answer:

The equation of the family of circles with radius 3 units, having its centre on y-axis, is given by
$x^{2} + {(y - a)}^{2} = 3^{2} . . . . . (1)$
Here, a is any arbitrary constant.
Since this equation has only one arbitrary constant, we get a first order differential equation.
Differentiating (1) with respect to x, we get
$2 x + 2 (y - a) \frac{d y}{d x} = 0 \Rightarrow x + (y - a) \frac{d y}{d x} = 0 \Rightarrow x = (a - y) \frac{d y}{d x} \Rightarrow \frac{x}{\frac{d y}{d x}} = a - y \Rightarrow a = y + \frac{x}{\frac{d y}{d x}}$
Substituting the value of a in (1), we get

$x^{2} + {(y - y - \frac{x}{\frac{d y}{d x}})}^{2} = 3^{2} \Rightarrow x^{2} + \frac{x^{2}}{{(\frac{d y}{d x})}^{2}} = 9 \Rightarrow x^{2} {(\frac{d y}{d x})}^{2} + x^{2} = 9 {(\frac{d y}{d x})}^{2} \Rightarrow x^{2} {(\frac{d y}{d x})}^{2} - 9 {(\frac{d y}{d x})}^{2} + x^{2} = 0 \Rightarrow (x^{2} - 9) {(\frac{d y}{d x})}^{2} + x^{2} = 0 \Rightarrow (x^{2} - 9) {(y')}^{2} + x^{2} = 0 Hence, (x^{2} - 9) {(y')}^{2} + x^{2} = 0 is the required differential equation.$

Page No 22.145:

Question 8:

Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Answer:

The equation of the parabola having vertex at origin and axis along the positive direction of y-axis is given by
x²=4ay .....(1)
Since there is only one parameter, so we differentiate it only once.
Differentiating with respect to x, we get
$2 x = 4 a y' \Rightarrow 4 a = \frac{2 x}{y'}$
Substituting the value of 4a in (1), we get
$x^{2} = \frac{2 x}{y'} \times y \Rightarrow x y' = 2 y \Rightarrow x y' - 2 y = 0$

Page No 22.145:

Question 9:

Form the differential equation of the family of ellipses having foci on y-axis and centre at the origin.

Answer:

The equation of the ellipses having foci on y-axis and centre at the origin is given by
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ .....(1)
Here,
b > a
Since these are two parameters, so we differentiate the equation twice.
Differentiating with respect to x, we get
$\frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} y' = 0 \Rightarrow \frac{x}{a^{2}} + \frac{y}{b^{2}} y' = 0 . . . . . (2) \Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} {(y')}^{2} + \frac{y}{b^{2}} y'' = 0 . . . . . (3) Multiplying throughout by x, we get \frac{x}{a^{2}} + \frac{x}{b^{2}} {(y')}^{2} + \frac{x y}{b^{2}} y'' = 0 . . . . . (4) Subtracting (2) from (4), we get \frac{1}{b^{2}} [x {(y')}^{2} + x y y'' - y y'] = 0 \Rightarrow x {(y')}^{2} + x y y'' - y y' = 0$

Page No 22.145:

Question 10:

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.

Answer:

The equation of the family of hyperbolas having centre at the origin and foci on the X-axis is given by
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 . . . . . (1)$
Here, a and b are parameters.
Since this equation contains two parameters, so we get a second order differential equation.
Differentiating (1) with respect to x, we get
$\frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} y' = 0 . . . . . (2)$
Differentiating (2) with respect to x, we get
$\frac{2}{a^{2}} - \frac{2}{b^{2}} [y y'' + {(y')}^{2}] = 0 \Rightarrow \frac{1}{a^{2}} = \frac{1}{b^{2}} [y y'' + {(y')}^{2}] \Rightarrow \frac{b^{2}}{a^{2}} = [y y'' + {(y')}^{2}] . . . . . (3)$
From (2), we get
$\frac{2 x}{a^{2}} = \frac{2 y}{b^{2}} y' \Rightarrow \frac{b^{2}}{a^{2}} = \frac{y}{x} y' . . . . . (4)$
From (3) and (4), we get
$\frac{y}{x} y' = [y y'' + {(y')}^{2}] \Rightarrow y y' = x y y'' + x {(y')}^{2} Hence, x y y'' + x {(y')}^{2} - y y' = 0 is the required differential equation.$

Page No 22.145:

Question 11:

Verify that xy = a e^x + b e^−x + x² is a solution of the differential equation $x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0$ .

Answer:

^{$We have, x y = a e^{x} + b e^{- x} + x^{2} Differentiating with respect to x on both sides, we get \Rightarrow x \frac{d y}{d x} + y = a e^{x} - b e^{- x} + 2 x Again differentiating with respect to x on both sides, we get \Rightarrow x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + \frac{d y}{d x} = a e^{x} + b e^{- x} + 2 \Rightarrow x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} = x y - x^{2} + 2 [∵ x y = a e^{x} + b e^{- x} + x^{2}] \Rightarrow x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0$}
Thus, xy = a e^x + b e^−x + x² is the solution of the given differential equation.

Page No 22.145:

Question 12:

Show that y = C x + 2C² is a solution of the differential equation $2 {(\frac{d y}{d x})}^{2} + x \frac{d y}{d x} - y = 0$ .

Answer:

We have,
$2 {(\frac{d y}{d x})}^{2} + x \frac{d y}{d x} - y = 0 . . . . . (1)$
Now,
y = C x + 2C²
^{$\Rightarrow \frac{d y}{d x} = C Putting \frac{d y}{d x} = C and y = C x + 2 C^{2} in (1), we get$}
$LHS = 2 {(C)}^{2} + x (C) - (C x + 2 C^{2}) = 2 C^{2} + x C - x C - 2 C^{2} = 0 = RHS$
Thus, y = C x + 2C² is the solution of the given differential equation.

Page No 22.145:

Question 13:

Show that y² − x² − xy = a is a solution of the differential equation $(x - 2 y) \frac{d y}{d x} + 2 x + y = 0$ .

Answer:

We have,
$(x - 2 y) \frac{d y}{d x} + 2 x + y = 0$
Now,y² − x² − xy = a
^{$\Rightarrow 2 y \frac{d y}{d x} - 2 x - y - x \frac{d y}{d x} = 0 \Rightarrow (2 y - x) \frac{d y}{d x} - 2 x - y = 0 \Rightarrow (2 y - x) \frac{d y}{d x} = 2 x + y \Rightarrow (x - 2 y) \frac{d y}{d x} = - (2 x + y) \Rightarrow (x - 2 y) \frac{d y}{d x} + 2 x + y = 0$}
Thus, y² − x² − xy = a is the solution of the given differential equation.

Page No 22.145:

Question 14:

Verify that y = A cos x + sin x satisfies the differential equation cos $x \frac{d y}{d x} + (\sin x)$ y = 1.

Answer:

We have,
$\cos x \frac{d y}{d x} + (\sin x) y = 1 . . . . . (1)$
Now,
y = A<