Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Given that $\large \frac{dy}{dx} = e^{-2y}$ and $y=0$ when $x=5$. Find the value of $x$ when $y=3$.

$\begin{array}{1 1}(A) \large\frac{e^6-\large9}{\large2} \\ (B) \large\frac{e^4-\large9}{\large2} \\(C) \large\frac{e^6-\large9}{\large3} \\ (D) \large\frac{2e^6-\large9}{\large2}\end{array}$

Can you answer this question?

1 Answer

0 votes
  • Form $\large\frac{dy}{dx}=f(x,y),$ when variables are seperable.
  • we expresss it in the form $f(x)dx=g(y) dy$
  • Then $\int g(y) dy=\int f(x) dx+c$
  • $\int e^{ax} dx=\large\frac{1}{a}$$e^{ax}+c$
Given $ \large\frac{dy}{dx}=e^{-2y}$ and y=0 when x=5
We have to find the value of x when y=3
Let us seperate the variables,
=>$ e^{2y}dy=dx$
On integrating both the sides we get,
$\int e^{2y}dy=\int dx$
When $y=0$ and $x=5$
$\large\frac{1}{2}$$ e^{2 \times 0}=5+c$
But $e^0=1$
Therefore $\large\frac{1}{2} $$=5+c$
When $y=3,$ let us find the value of x
$\large\frac{1}{2} $$e^{2 \times 3}=x+\large\frac{9}{2}$
Hence the value is $\Large\frac{e^6-\large9}{\large2}$
answered May 7, 2013 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App