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#### Question 1:

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

We have,

#### Question 4:

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

#### Question 5:

$x\frac{dy}{dx}=x+y$

#### Question 6:

$\frac{dy}{dx}+2y=4x$

#### Question 8:

$\frac{dy}{dx}+\frac{4x}{{x}^{2}+1}y+\frac{1}{{\left({x}^{2}+1\right)}^{2}}=0$

#### Question 9:

We have,

Dividing both sides by x, we get

#### Question 11:

$\frac{dy}{dx}+\frac{y}{x}={x}^{3}$

#### Question 15:

$\frac{dy}{dx}$ = y tan x − 2 sin x

#### Question 17:

$\frac{dy}{dx}$ + y tan x = cos x

#### Question 18:

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

We have,

#### Question 21:

x dy = (2y + 2x4 + x2) dx

#### Question 22:

$\left(1+{y}^{2}\right)+\left(x-{e}^{{\mathrm{tan}}^{-1}y}\right)\frac{dy}{dx}=0$

#### Question 23:

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

#### Question 24:

$\left(2x-10{y}^{3}\right)\frac{dy}{dx}+y=0$

#### Question 25:

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

#### Question 26:

dx + xdy = ey sec2 y dy

#### Question 27:

$\frac{dy}{dx}$ = y tan x − 2 sin x

#### Question 28:

$\frac{dy}{dx}$ + y cos x = sin x cos x

#### Question 29:

Solve the following differential equations:

$\left(1+{x}^{2}\right)\frac{dy}{dx}-2xy=\left({x}^{2}+2\right)\left({x}^{2}+1\right)$        [CBSE 2005]

Given, $\left(1+{x}^{2}\right)\frac{dy}{dx}-2xy=\left({x}^{2}+2\right)\left({x}^{2}+1\right)$
$⇒\frac{dy}{dx}-\frac{2x}{\left(1+{x}^{2}\right)}y=\left({x}^{2}+2\right)\phantom{\rule{0ex}{0ex}}$
This is a linear differential equation.
I.F.$={e}^{\int -\frac{2x}{1+{x}^{2}}dx}=\frac{1}{1+{x}^{2}}$
$y\left(\frac{1}{1+{x}^{2}}\right)=\int \left(\frac{{x}^{2}+2}{{x}^{2}+1}\right)dx\phantom{\rule{0ex}{0ex}}⇒y\left(\frac{1}{1+{x}^{2}}\right)=\int \left(1+\frac{1}{1+{x}^{2}}\right)dx\phantom{\rule{0ex}{0ex}}⇒y\left(\frac{1}{1+{x}^{2}}\right)=x+{\mathrm{tan}}^{-1}x+C\phantom{\rule{0ex}{0ex}}⇒y=\left(x+{\mathrm{tan}}^{-1}x+C\right)\left(1+{x}^{2}\right)$

#### Question 31:

$\left({x}^{2}-1\right)\frac{dy}{dx}+2\left(x+2\right)y=2\left(x+1\right)$

#### Question 33:

$\frac{dy}{dx}-y=x{e}^{x}$

#### Question 34:

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

#### Question 35:

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

#### Question 36:

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

(ii)

(iii) $x\frac{dy}{dx}-y=\left(x+1\right){e}^{-x}$

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

(v)

(vi) $\frac{dy}{dx}-\frac{2xy}{1+{x}^{2}}={x}^{2}+2$

(vii)

(viii) $\left(x+y\right)\frac{dy}{dx}=1$

(ix)

(x)

(xi)

(xii)

#### Question 37:

Solve each of the following initial value problems:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)
(xi)
(xii) $dy=\mathrm{cos}x\left(2-y\mathrm{cos}ecx\right)dx$
(xiii) given that y = 0 when $x=\frac{\mathrm{\pi }}{2}$.

(xiii)
$\mathrm{tan}x\left(\frac{dy}{dx}\right)=2x\mathrm{tan}x+{x}^{2}-y\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}+\frac{1}{\mathrm{tan}x}y=\frac{2x\mathrm{tan}x+{x}^{2}}{\mathrm{tan}x}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}+\left(\mathrm{cot}x\right)y=2x+{x}^{2}\mathrm{cot}x$
This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$.

Integrating factor, I.F. = ${e}^{\int Pdx}={e}^{\int \mathrm{cot}xdx}={e}^{\mathrm{log}\mathrm{sin}x}=\mathrm{sin}x$

The solution of the given differential equation is given by

$y×\left(\mathrm{I}.\mathrm{F}.\right)=\int Q×\left(\mathrm{I}.\mathrm{F}.\right)dx+C\phantom{\rule{0ex}{0ex}}⇒y×\mathrm{sin}x=\int \left(2x+{x}^{2}\mathrm{cot}x\right)\mathrm{sin}xdx+C\phantom{\rule{0ex}{0ex}}⇒y\mathrm{sin}x=\int 2x\mathrm{sin}xdx+\int {x}^{2}\mathrm{cos}xdx+C\phantom{\rule{0ex}{0ex}}⇒y\mathrm{sin}x=\int 2x\mathrm{sin}xdx+\left[{x}^{2}\int \mathrm{cos}xdx-\int \left(\frac{d}{dx}{x}^{2}×\int \mathrm{cos}xdx\right)dx\right]+C$

It is given that, y = 0 when $x=\frac{\mathrm{\pi }}{2}$.

$\therefore 0={\left(\frac{\mathrm{\pi }}{2}\right)}^{2}+\mathrm{cosec}\frac{\mathrm{\pi }}{2}×C\phantom{\rule{0ex}{0ex}}⇒C=-\frac{{\mathrm{\pi }}^{2}}{4}$
Putting $C=-\frac{{\mathrm{\pi }}^{2}}{4}$ in (1), we get

$y={x}^{2}-\frac{{\mathrm{\pi }}^{2}}{4}\mathrm{cosec}x$

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

#### Question 38:

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

#### Question 39:

Find the general solution of the differential equation .

#### Question 40:

Solve the differential equation $\left(y+3{x}^{2}\right)\frac{dx}{dy}=x$

#### Question 41:

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

#### Question 42:

Solve the following differential equation:

The given differential equation is .
This differential equation can be written as

This is a linear differential equation with .
I.F. =  ${\mathrm{e}}^{-\int \frac{1}{1+{y}^{2}}dy}={\mathrm{e}}^{{\mathrm{cot}}^{-1}y}$
Multiply the differential equation by integration factor (I.F.), we get

Integrating both sides with respect y, we get

Putting $t={\mathrm{cot}}^{-1}y$ and $dt=-\frac{1}{1+{y}^{2}}dy$, we get

#### 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.

Let r be the radius and S be the surface area of the balloon at any time t. Then,

#### Question 2:

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

Let P0 be the initial population and P be the population at any time t. Then,
$\frac{dP}{dt}=\frac{5P}{100}\phantom{\rule{0ex}{0ex}}⇒\frac{dP}{dt}=0.05P$

#### 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?

Let the original population be N and the population at any time t be P.
Given: $\frac{dP}{dt}\alpha P$

#### 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?

#### 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?

Let P0 be the initial amount and P be the amount at any time t. Then,
$\frac{dP}{dt}=\frac{6P}{100}\phantom{\rule{0ex}{0ex}}⇒\frac{dP}{dt}=0.06P$

#### 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.

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: $\frac{dP}{dt}\alpha P$

#### 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?

Let the population at any time t be P.

Given:

#### Question 8:

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

#### 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.

Let P0 be the initial amount and P be the amount at any time t.
We have,
$\frac{dP}{dt}=\frac{8P}{100}\phantom{\rule{0ex}{0ex}}⇒\frac{dP}{dt}=\frac{2P}{25}$

#### 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 $\frac{di}{dt}$+ R i = E. If E is constant and initially no current passes through the circuit, prove that $i=\frac{E}{R}\left\{1-{e}^{-\left(R/L\right)t}\right\}$.

#### 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.

Let the initial amount of radium be N and the amount of radium present at any time t be P.
Given: $\frac{dP}{dt}\alpha P$

#### 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?

Let the original amount of the radium be N and the amount of radium at any time t be P.
Given: $\frac{dP}{dt}\alpha P$

#### 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.

#### Question 14:

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

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

$⇒\frac{dy}{dx}\left(x+1\right)=y\left(1-y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{y\left(1-y\right)}=\frac{dx}{\left(x+1\right)}\phantom{\rule{0ex}{0ex}}$

#### Question 15:

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

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

#### 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.

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
, where (X, Y) is the arbitrary point on the tangent.
Putting Y=0 we get,

Integrating on both sides we get,

#### 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) = 2e2x.

#### 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).

The slope of the curve is given as .
Here,

#### 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).

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

It is given, OT = 2x

$\begin{array}{l}\therefore x-y \frac{dx}{dy}=\mathrm{2}x\\ -x=y\frac{dx}{dy}\\ -\int \frac{dx}{x}=\int \frac{dy}{y}\\ \therefore xy=k\end{array}$

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

#### Question 20:

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

#### Question 21:

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

According to the question,
$\frac{dy}{dx}=\frac{2y}{x}$

#### 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.

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

$⇒\frac{dy}{dx}-3y=x-1$

#### 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.

According to the question,

$\frac{dy}{dx}=x+xy\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=x\left(1+y\right)$

#### 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 ${y}^{2}-2xy\frac{dy}{dx}-{x}^{2}=0$, and hence find the curve.

Tangent  at P(x, y) is given by
If p be the perpendicular from the origin, then

Multiplying by the integrating factor ${e}^{\int -\frac{1}{x}dx}=\frac{1}{x}$

#### 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.

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

#### 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.

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
$Y-y=-\frac{1}{\frac{dy}{dx}}\left(X-x\right)$
It is given that the curve passes through the point (3, 0). Then,

#### 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.

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: $\frac{dP}{dt}\alpha P$

#### 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?

Let the original amount of radium be N and the amount of radium at any time t be P.
Given: $\frac{dP}{dt}\alpha P$

#### Question 29:

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

We have,

Thus, ${x}^{2}-{y}^{2}=Cx$ is the equation of the rectangular hyperbola.

#### 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.

According to the question,
$\frac{dy}{dx}=x+y$
$⇒\frac{dy}{dx}-y=x$

#### 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.

According to the question,
$\frac{dy}{dx}=x+xy$
$⇒\frac{dy}{dx}-xy=x$

#### 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).

According to the question,
$\frac{dy}{dx}={x}^{2}$

#### 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.

According to the question,

#### 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).

Let P(x, y) be any point on the curve. Then slope of the tangent at P is $\frac{dy}{dx}$.
It is given that the slope of the tangent at P(x,y) is equal to the ordinate  i.e y.
Therefore $\frac{dy}{dx}$ = y

Since, the curve passes through (1,1). Therefore, x=1 and y=1 .
Putting these values in equation obtained above we get,

#### Question 1:

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

(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 + ey' = 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 + ey' = 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.

#### Question 2:

Verify that the function y = e−3x is a solution of the differential equation .

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

#### 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 = ex + 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' = $\frac{xy}{1+{x}^{2}}$ (v) y = x sin x xy' = y + x $\sqrt{{x}^{2}-{y}^{2}}$ (vi) $y=\sqrt{{a}^{2}-{x}^{2}}$ $x+y\frac{dy}{dx}=0$

(i) We have,
y'' − y' = 0                            .....(1)
Now,
y = ex +1

$⇒y\text{'}={e}^{x}\phantom{\rule{0ex}{0ex}}⇒y\text{'}\text{'}={e}^{x}$

Putting the above values in (1), we get

Thus, y = ex +1 is the solution of the given differential equation.

(ii) We have,
y' − 2x − 2 = 0                    .....(1)
Now,
y = x2 + 2x + C
$⇒y\text{'}=2x+2$
Putting the above value in (1), we get
$\mathrm{LHS}=2x+2-2x-2=0=\mathrm{RHS}$
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

Putting the above value in (1), we get

Thus, y = cos x + C is the solution of the given differential equation.

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

Putting the above value in (1), we get
$\mathrm{LHS}=\frac{x}{\sqrt{1+{x}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{x}{\sqrt{1+{x}^{2}}}×\frac{\sqrt{1+{x}^{2}}}{\sqrt{1+{x}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{xy}{1+{x}^{2}}=\mathrm{RHS}$
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

Putting the above value in (1), we get

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

Putting the above value in (1), we get

Thus, y = x sin x is the solution of the given differential equation.

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

Now,
$y=\sqrt{{a}^{2}-{x}^{2}}$
$⇒y\text{'}=\frac{-x}{\sqrt{{a}^{2}-{x}^{2}}}\phantom{\rule{0ex}{0ex}}$

Putting the above value in (1), we get

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

#### Question 4:

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

We have,
y = mx          (1)
Differentiating both sides, we get

#### Question 5:

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

We have,
y = a sin (x + b)          .....(2)
Differentiating both sides, we get

#### Question 6:

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

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
$2y\frac{dy}{dx}=4a$
Substituting the value of 4a in (1), we get
${y}^{2}=2y\frac{dy}{dx}×x\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=2xy\frac{dy}{dx}\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-2xy\frac{dy}{dx}=0\phantom{\rule{0ex}{0ex}}$

#### Question 7:

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

The equation of the family of circles with radius 3 units, having its centre on y-axis, is given by

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

Substituting the value of a in (1), we get

#### Question 8:

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

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
$2x=4ay\text{'}\phantom{\rule{0ex}{0ex}}⇒4a=\frac{2x}{y\text{'}}$
Substituting the value of 4a in (1), we get
${x}^{2}=\frac{2x}{y\text{'}}×y\phantom{\rule{0ex}{0ex}}⇒xy\text{'}=2y\phantom{\rule{0ex}{0ex}}⇒xy\text{'}-2y=0\phantom{\rule{0ex}{0ex}}$

#### Question 9:

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

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

#### Question 10:

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

The equation of the family of hyperbolas having centre at the origin and foci on the X-axis is given by

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

Differentiating (2) with respect to x, we get

From (2), we get

From (3) and (4), we get

#### Question 11:

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

Thus, xy = a ex + b ex + x2 is the solution of the given differential equation.

#### Question 12:

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

We have,

Now,
y = C x + 2C2

Thus, y = C x + 2C2 is the solution of the given differential equation.

#### Question 13:

Show that y2x2xy = a is a solution of the differential equation $\left(x-2y\right)\frac{dy}{dx}+2x+y=0$.

We have,
$\left(x-2y\right)\frac{dy}{dx}+2x+y=0$
Now,y2x2xy = a
$⇒2y\frac{dy}{dx}-2x-y-x\frac{dy}{dx}=0\phantom{\rule{0ex}{0ex}}⇒\left(2y-x\right)\frac{dy}{dx}-2x-y=0\phantom{\rule{0ex}{0ex}}⇒\left(2y-x\right)\frac{dy}{dx}=2x+y\phantom{\rule{0ex}{0ex}}⇒\left(x-2y\right)\frac{dy}{dx}=-\left(2x+y\right)\phantom{\rule{0ex}{0ex}}⇒\left(x-2y\right)\frac{dy}{dx}+2x+y=0$
Thus, y2x2xy = a is the solution of the given differential equation.

#### Question 14:

Verify that y = A cos x + sin x satisfies the differential equation cos y = 1.

We have,

Now,
y = A cos x + sin x

Thus, y = A cos x + sin x is the solution of the given differential equation.

#### Question 15:

Find the differential equation corresponding to y = ae2x + be3x + cex where a, b, c are arbitrary constants.

We have,
y
= ae2x + be3x + cex            .....(1)
Differentiating with respect to x, we get

#### Question 16:

Show that the differential equation of all parabolas which have their axes parallel to y-axis is $\frac{{d}^{3}y}{d{x}^{3}}=0$.

The equation of the family of parabolas axis parallel to y-axis is given by
${\left(x-\beta \right)}^{2}=4a\left(y-\alpha \right)$                                         .....(1)
Here,  are two arbitrary constants.

Differentiating (1) with respect to x, we get
$2\left(x-\beta \right)=4a\frac{dy}{dx}\phantom{\rule{0ex}{0ex}}⇒1=2a\frac{{d}^{2}y}{d{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒0=2a\frac{{d}^{3}y}{d{x}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{d}^{3}y}{d{x}^{3}}=0$

#### Question 17:

From x2 + y2 + 2ax + 2by + c = 0, derive a differential equation not containing a, b and c.

We have,
x2 + y2 + 2ax + 2by + c = 0         .....(i)

Differentiating (i) with respect to x, we get

#### Question 19:

$\frac{dy}{dx}=\frac{1}{{x}^{2}+4x+5}$

#### Question 20:

$\frac{dy}{dx}={y}^{2}+2y+2$

#### Question 21:

$\frac{dy}{dx}+4x={e}^{x}$

#### Question 24:

(tan2 x + 2 tan x + 5) $\frac{dy}{dx}=2$ (1 + tan x) sec2 x

#### Question 26:

tan y dx + tan x dy = 0

We have,
tan y dx + tan x dy = 0

#### Question 27:

(1 + x) y dx + (1 + y) x dy = 0

We have,
(1 + x) y dx + (1 + y) x dy = 0

#### Question 28:

x cos2 y dx = y cos2 x dy

We have,
x cos2 y dx = y cos2 x dy

#### Question 29:

cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy

We have,
cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy

#### Question 30:

cosec x (log y) dy + x2y dx = 0

We have,

#### Question 31:

(1 − x2) dy + xy dx = xy2 dx

We have,

We have,

#### Question 33:

x (e2y − 1) dy + (x2 − 1) ey dx = 0

#### Question 34:

$\frac{dy}{dx}+1={e}^{x+y}$

We have,

#### Question 35:

$\frac{dy}{dx}={\left(x+y\right)}^{2}$

#### Question 36:

cos (x + y) dy = dx

#### Question 37:

$\frac{dy}{dx}+\frac{y}{x}=\frac{{y}^{2}}{{x}^{2}}$

#### Question 38:

$\frac{dy}{dx}=\frac{y\left(x-y\right)}{x\left(x+y\right)}$

#### Question 39:

(x + y − 1) dy = (x + y) dx

#### Question 44:

(1 + y + x2 y) dx + (x + x3) dy = 0

On integrating both side we get,

$\left(xy\right)=-\int \frac{1}{1+{x}^{2}}dx\phantom{\rule{0ex}{0ex}}⇒xy=-{\mathrm{tan}}^{-1}x+c\phantom{\rule{0ex}{0ex}}⇒xy+{\mathrm{tan}}^{-1}x=c$

#### Question 45:

(x2 + 1) dy + (2y − 1) dx = 0

#### Question 46:

y sec2 x + (y + 7) tan x $\frac{dy}{dx}$ = 0

#### Question 47:

(2ax + x2) $\frac{dy}{dx}$ = a2 + 2ax

#### Question 48:

(x3 − 2y3) dx + 3x2 y dy = 0

Disclaimer: There seems to be error in the given question.

#### Question 49:

x2 dy + (x2xy + y2) dx = 0

#### Question 50:

$y-x\frac{dy}{dx}=b\left(1+{x}^{2}\frac{dy}{dx}\right)$

#### Question 52:

$\frac{dy}{dx}+y=4x$

#### Question 56:

x cos x $\frac{dy}{dx}$ + y (x sin x + cos x) = 1

We have,

#### Question 58:

${y}^{2}+\left(x+\frac{1}{y}\right)\frac{dy}{dx}=0$

We have,
${y}^{2}+\left(x+\frac{1}{y}\right)\frac{dy}{dx}=0$
$⇒\frac{dy}{dx}=-\frac{{y}^{3}}{xy+1}\phantom{\rule{0ex}{0ex}}⇒\frac{dx}{dy}=-\frac{xy+1}{{y}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{dx}{dy}=-\frac{x}{{y}^{2}}-\frac{1}{{y}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{dx}{dy}+\frac{x}{{y}^{2}}=-\frac{1}{{y}^{3}}$

#### Question 59:

2 cos $x\frac{dy}{dx}+4y$ sin x = sin 2x, given that y = 0 when x = $\frac{\mathrm{\pi }}{3}$.

We have,

#### Question 60:

(1 + y2) dx = (tan−1 yx) dy

#### Question 61:

$\frac{dy}{dx}$ + y tan x = xn cos x, n ≠ − 1

#### Question 62:

Find the general solution of the differential equation .

#### Question 63:

Find the particular solution of the differential equation $\frac{dy}{dx}=-4x{y}^{2}$ given that y = 1, when x = 0.

#### Question 64:

For each of the following differential equations, find the general solution:
(i)

(ii)

(iii) $\frac{dy}{dx}=\left(1+{x}^{2}\right)\left(1+{y}^{2}\right)$

(iv) y log y dxx dy = 0

(v) $\frac{dy}{dx}={\mathrm{sin}}^{-1}x$

(vi) $\frac{dy}{dx}+y=1$

#### Question 65:

For each of the following differential equations, find a particular solution satisfying the given condition:
(i)

(ii)

(iii)

#### Question 66:

Solve the each of the following differential equations:
(i) $\left(x-y\right)\frac{dy}{dx}=x+2y$

(ii)

(iii) y dx + x log $\left(\frac{y}{x}\right)$ dy − 2x dy = 0

(iv)

(v)

(vi)

(vii) $\frac{dy}{dx}+3y={e}^{-2x}$

(viii) $\frac{dy}{dx}+\frac{y}{x}={x}^{2}$

(ix)

(x)

(xi)

(xii) (1 + x2) dy + 2xy dx = cot x dx

(xiii) $\left(x+y\right)\frac{dy}{dx}=1$

(xiv) y dx + (xy2) dy = 0

(xv) $\left(x+3{y}^{2}\right)\frac{dy}{dx}=y$