# Let y=y(x) be the solution of the differential equation

Question:

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 $\mathrm{y}(2)=0$. Then the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=1$ is equal to:

1. (1) $\frac{-\mathrm{e}^{3 / 2}}{\left(\mathrm{e}^{2}+1\right)^{2}}$

2. (2) $-\frac{2 \mathrm{e}^{2}}{\left(1+\mathrm{e}^{2}\right)^{2}}$

3. (3) $\frac{\mathrm{e}^{5 / 2}}{\left(1+\mathrm{e}^{2}\right)^{2}}$

4. (4) $\frac{5 \mathrm{e}^{1 / 2}}{\left(\mathrm{e}^{2}+1\right)^{2}}$

Correct Option: 1

Solution:

Let $\mathbf{y}+1=\mathrm{Y}$

$\therefore \frac{\mathrm{d} \mathrm{Y}}{\mathrm{dx}}=\mathrm{Y}^{2} \mathrm{e}^{\frac{x^{2}}{2}}-\mathrm{x} \mathrm{Y}$

Put $-\frac{1}{Y}=k$

$\Rightarrow \frac{\mathrm{dk}}{\mathrm{dx}}+\mathrm{k}(-\mathrm{x})=\mathrm{e}^{\frac{\mathrm{x}^{2}}{2}}$

I.F. $=\mathrm{e}^{-\frac{x^{2}}{2}}$

$\therefore \mathrm{k}=(\mathrm{x}+\mathrm{c}) \mathrm{e}^{\mathrm{x}^{2} / 2}$

Put $k=-\frac{1}{y+1}$

$\therefore y+1=-\frac{1}{(x+c) e^{x^{2} / 2}}$

when $x=2, y=0$, then $c=-2-\frac{1}{e^{2}}$

diffentiate equation

(i) \& put $x=1$

we get $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=1}=-\frac{\mathrm{e}^{3 / 2}}{\left(1+\mathrm{e}^{2}\right)^{2}}$