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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Given that $\large \frac{dy}{dx} = e^{-2y}$ and $y=0$ when $x=5$. Find the value of $x$ when $y=3$.

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

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1 Answer

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Toolbox:
  • 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}$$e^{ax}+c$
Given $ \large\frac{dy}{dx}=e^{-2y}$ and y=0 when x=5
We have to find the value of x when y=3
Let us seperate the variables,
$\Large\frac{dy}{e^{-2y}}$$=dx$
=>$ e^{2y}dy=dx$
On integrating both the sides we get,
$\int e^{2y}dy=\int dx$
=>$\large\frac{1}{2}$$e^{2y}=x+c$
When $y=0$ and $x=5$
$\large\frac{1}{2}$$ e^{2 \times 0}=5+c$
But $e^0=1$
Therefore $\large\frac{1}{2} $$=5+c$
$-c=\large\frac{9}{2}$
When $y=3,$ let us find the value of x
$\large\frac{1}{2} $$e^{2 \times 3}=x+\large\frac{9}{2}$
$=\Large\frac{e^6}{2}-\large\frac{9}{2}=$$x$
$=\Large\frac{e^6-\large9}{\large2}$
Hence the value is $\Large\frac{e^6-\large9}{\large2}$
answered May 7, 2013 by meena.p
 
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