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Shanaya
Subject: Maths
, asked on 18/3/18
In the second step how did we get 4 as power In RHS?
Answer
1
Archit
Subject: Maths
, asked on 18/3/18
Find the differential eq. .. As the general solution
Q.11. Which of the following differential equations has
$y={c}_{1}{e}^{x}+{c}_{2}{e}^{x}$
as the general solution?
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 18/3/18
Q If
$y={e}^{a{\mathrm{cos}}^{1}x},thenshowthat(1{x}^{2}){y}_{2}x{y}_{1}{a}^{2}y=0$
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 17/3/18
No links please
Q. iii) Solve :
$\left(x{y}^{2}{e}^{\frac{1}{{x}^{3}}}\right)dx{x}^{2}ydy=0$
; given y = 0 when x = 1.
Ans:
${y}^{2}=\frac{2}{2}\left({e}^{{x}^{\frac{1}{3}}}e\right){x}^{2}$
Answer
1
Manasi Mujumdar
Subject: Maths
, asked on 16/3/18
Solve the following
Differential equation

$\left(y+3{x}^{2}\right)\frac{dx}{dy}=x$
Answer
1
Bhanvi
Subject: Maths
, asked on 16/3/18
In a college hostel accommodating 1000 students, one of the hostellers came in carrying a flu virus, and the hostel was isolated. If the rate at which the virus spreads is assumed to be proportional to the product of the number N of infected students and the number of noninfected students, and if infected students are 50 after 4 days then show that more than 95% of the hostellers will be infected after 10 days. If Shyam was the first student to be infected what precautions he should have initiated to avoid this situation.
Answer
1
Aishwarya Trivedi
Subject: Maths
, asked on 15/3/18
in this soln part i didnt get after (((log(1y)=x+logc)))) . in the next line why there is a minus sign before log(1y)??? because when logc is bring to the left side it should be ve but why log(1y)is ve?:
Now, Integrating both sides, we Get:
$\int \frac{dy}{1y}=\int dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}\left(1y\right)=x+\mathrm{log}C\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}C\mathrm{log}\left(1y\right)=x\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}C\left(1y\right)=x\phantom{\rule{0ex}{0ex}}\Rightarrow C(1y)={e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow 1y=\frac{1}{c}{e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow y=1\frac{1}{c}{e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow y=A{e}^{x}\left(WhereA=\frac{{\displaystyle 1}}{{\displaystyle c}}\right)$
$\int \frac{dy}{1y}=\int dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}\left(1y\right)=x+\mathrm{log}C\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}C\mathrm{log}\left(1y\right)=x\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}C\left(1y\right)=x\phantom{\rule{0ex}{0ex}}\Rightarrow C(1y)={e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow 1y=\frac{1}{c}{e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow y=1\frac{1}{c}{e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow y=A{e}^{x}\left(WhereA=\frac{{\displaystyle 1}}{{\displaystyle c}}\right)$
Answer
1
Jeneeta Eliza John
Subject: Maths
, asked on 15/3/18
$\sqrt{a+x}\frac{dy}{dx}+xy=0$
Answer
2
Megha Prasadan
Subject: Maths
, asked on 15/3/18
Q. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Answer
1
Devi Das
Subject: Maths
, asked on 14/3/18
Q). Solve:
$\frac{dy}{dx}=\frac{2x3y+4}{3x+4y5}$
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 13/3/18
No links please
$\left(i\right)Showthat\left[\overrightarrow{a}\times \overrightarrow{b}\overrightarrow{b}\times \overrightarrow{c}\overrightarrow{c}\times \overrightarrow{a}\right]{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}^{2}$
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 13/3/18
No links please. Please explain.
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 13/3/18
No links please
$\left(iii\right)Solve:x\frac{dy}{dx}y=\mathrm{log}x,wheny\left(1\right)=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(iii\right)Ans:\frac{y}{x}+\frac{1}{x}(\mathrm{log}x+1)=1$
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 13/3/18
No links please
$Ify=\mathrm{cos}\left(m{\mathrm{sin}}^{1}x\right),thenshowthat\left(1{x}^{2}\right)\frac{{d}^{2}y}{d{x}^{2}}x\frac{dy}{dx}+{m}^{2}y=0.$
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 13/3/18
No links please
Q(i). If y = tan ( x + y ), then show that
$\frac{{d}^{2}y}{d{x}^{2}}=\frac{2\left(1+{y}^{2}\right)}{{y}^{5}}$
No links please
Answer
2
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What are you looking for?
Q.11. Which of the following differential equations has $y={c}_{1}{e}^{x}+{c}_{2}{e}^{x}$ as the general solution?
Q. iii) Solve : $\left(x{y}^{2}{e}^{\frac{1}{{x}^{3}}}\right)dx{x}^{2}ydy=0$; given y = 0 when x = 1.
Ans: ${y}^{2}=\frac{2}{2}\left({e}^{{x}^{\frac{1}{3}}}e\right){x}^{2}$
$\left(y+3{x}^{2}\right)\frac{dx}{dy}=x$
Now, Integrating both sides, we Get:
$\int \frac{dy}{1y}=\int dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}\left(1y\right)=x+\mathrm{log}C\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}C\mathrm{log}\left(1y\right)=x\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}C\left(1y\right)=x\phantom{\rule{0ex}{0ex}}\Rightarrow C(1y)={e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow 1y=\frac{1}{c}{e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow y=1\frac{1}{c}{e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow y=A{e}^{x}\left(WhereA=\frac{{\displaystyle 1}}{{\displaystyle c}}\right)$$\int \frac{dy}{1y}=\int dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}\left(1y\right)=x+\mathrm{log}C\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}C\mathrm{log}\left(1y\right)=x\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}C\left(1y\right)=x\phantom{\rule{0ex}{0ex}}\Rightarrow C(1y)={e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow 1y=\frac{1}{c}{e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow y=1\frac{1}{c}{e}^{x}\phantom{\rule{0ex}{0ex}}\Rightarrow y=A{e}^{x}\left(WhereA=\frac{{\displaystyle 1}}{{\displaystyle c}}\right)$
$\left(i\right)Showthat\left[\overrightarrow{a}\times \overrightarrow{b}\overrightarrow{b}\times \overrightarrow{c}\overrightarrow{c}\times \overrightarrow{a}\right]{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}^{2}$
$\left(iii\right)Solve:x\frac{dy}{dx}y=\mathrm{log}x,wheny\left(1\right)=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(iii\right)Ans:\frac{y}{x}+\frac{1}{x}(\mathrm{log}x+1)=1$
$Ify=\mathrm{cos}\left(m{\mathrm{sin}}^{1}x\right),thenshowthat\left(1{x}^{2}\right)\frac{{d}^{2}y}{d{x}^{2}}x\frac{dy}{dx}+{m}^{2}y=0.$
Q(i). If y = tan ( x + y ), then show that $\frac{{d}^{2}y}{d{x}^{2}}=\frac{2\left(1+{y}^{2}\right)}{{y}^{5}}$ No links please