JEE Main 2023 — Differential Equations Question with Solution

Given,

$\left(1-x^2{y}^2\right)dx=ydx+xdy$

$\Rightarrow dx=\frac{ydx+xdy}{1-{\left(xy\right)}^2}$

$\Rightarrow dx=\frac{d\left(yx\right)}{1-{\left(xy\right)}^2}$

$\Rightarrow dx=\frac{1}{2}\left(\frac{d\left(xy\right)}{1-xy}+\frac{d\left(xy\right)}{1+xy}\right)$

Now integrating both side we get,

$\Rightarrow 2x+c=\ln \left|\frac{1+xy}{1-xy}\right|$

$\Rightarrow \left|\frac{xy+1}{xy-1}\right|=e^c{e}^{2x}$

Now given, $y\left(1\right)=2$ so putting the value in above equation we get,

$3=e^c{e}^2$

$\Rightarrow e^c=\frac{3}{e^2}$

Hence, the equation becomes $\left|\frac{xy+1}{xy-1}\right|=3e^{2x-2}$

Now finding $y\left(2\right)$ so putting $x=2$ in above equation we get, 

$\left|\frac{2y+1}{2y-1}\right|=3e^2$

$\Rightarrow 2y+1=2·3e^{2}y-3e^2$

$\Rightarrow 1+3e^2=2y\left(3e^2-1\right)$

$\Rightarrow y\left(2\right)=\frac{1+3e^2}{2\left(3e^2-1\right)}$