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

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

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