Let $y=y(x)$ be the solution of the differential
equation $\frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{y}+1)\left((\mathrm{y}+1) \mathrm{e}^{\mathrm{x}^{2} / 2}-\mathrm{x}\right), 0<\mathrm{x}<2.1$
with $y(2)=0$. Then the value of $\frac{d y}{d x}$ at
$x=1$ is equal to :
Correct Option: 1
Let $\mathrm{y}+1=\mathrm{Y}$
$\therefore \frac{\mathrm{dY}}{\mathrm{dx}}=\mathrm{Y}^{2} \mathrm{e}^{\frac{\mathrm{x}^{2}}{2}}-\mathrm{XY}$
Put $-\frac{1}{Y}=k$
$\Rightarrow \frac{\mathrm{dk}}{\mathrm{dx}}+\mathrm{k}(-\mathrm{x})=\mathrm{e}^{\frac{\mathrm{x}^{2}}{2}}$
I.F. $=e^{-\frac{x^{2}}{2}}$
$\therefore \mathrm{k}=(\mathrm{x}+\mathrm{c}) \mathrm{e}^{\mathrm{x}^{2} / 2}$
Put $k=-\frac{1}{y+1}$
$\therefore \mathrm{y}+1=-\frac{1}{(\mathrm{x}+\mathrm{c}) \mathrm{e}^{\mathrm{x}^{2} / 2}}$ ........(1)
when $x=2, y=0$, then $c=-2-\frac{1}{e^{2}}$
diffentiate equation (i) \& put $x=1$
we get $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=1}=-\frac{\mathrm{e}^{3 / 2}}{\left(1+\mathrm{e}^{2}\right)^{2}}$
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