Question

Let $y=y\left(x\right)$ be the solution of the differential equation $\frac{dy}{dx}=(y+1)\left(\right(y+1)e^{\frac{x^2}{2}}-x),0<x<2,$ such that $y\left(2\right)=0.$ Then the value of $\frac{dy}{dx}$ at $x=1$ is

• A. $-\frac{e^{3/2}}{(e^2+1{)}^2}$
• B. $-\frac{e^{1/2}}{(e^2+1)}$
• C. $\frac{e^{1/2}}{(e^2+1{)}^2}(e^{1/2}-1)$
• D. $\frac{e^{3/2}}{(e^2+1{)}^2}(e^{3/2}+1)$

Answer 1

The correct option is A $-\frac{e^{3/2}}{(e^2+1{)}^2}$

$\frac{dy}{dx}=(y+1)\left(\right(y+1)e^{\frac{x^2}{2}}-x)\cdots \left(i\right)$
$\Rightarrow \frac{-1}{(y+1{)}^2}\frac{dy}{dx}-x\left(\frac{1}{y+1}\right)=-e^{x^2/2}$
Let $\frac{1}{1+y}=z\Rightarrow -\frac{1}{(1+y{)}^2}dy=dz$
$\Rightarrow \frac{dz}{dx}-xz=-e^{x^2/2}$

$I.F.=e^{\int -xdx}=e^{-x^2/2}$
$\Rightarrow z\left(e^{-x^2/2}\right)=-\int e^{-x^2/2}\cdot e^{x^2/2}dx$
$\Rightarrow \frac{1}{y+1}\left(e^{-x^2/2}\right)=-x+c$
as $y\left(2\right)=0$
$\therefore c=e^{-2}+2$
$\Rightarrow y+1=\frac{e^{-x^2/2}}{e^{-2}+2-x}$

At $x=1,y+1=\frac{e^{3/2}}{e^2+1}$
Using $\left(i\right)$
${\frac{dy}{dx}|}_{x=1}=\frac{e^{3/2}}{e^2+1}(\frac{e^{3/2}}{e^2+1}\cdot e^{1/2}-1)$
$=-\frac{e^{3/2}}{(e^2+1{)}^2}$