JEE Main 2023 — Differential Equations Question with Solution

Given,

$2\left(y+2\right){\log }_e\left(y+2\right)dx+\left(x+4-2{\log }_e\left(y+2\right)\right)dy=0$

Let $x+4=u, y+2=v$

$dx=du, dy=dv$

So, the equation becomes,

$\left(2v \ln v\right)\mathrm{du}=-\left(u-2 \ln v\right)dv$

$\Rightarrow 2v \ln v\frac{du}{dv}+u=2 \ln v$

$\Rightarrow \frac{du}{dv}+\frac{1}{2v \ln v}. u=\frac{1}{v}$

Which is a linear differential equation,

So, $IF=e^{\frac{1}{2}\int \frac{1}{v \ln v}}=e^{\frac{1}{2}\ln \left(\ln v\right)}={\left(\ln v\right)}^{\frac{1}{2}}$

Now solution of differential equation is given by,

$u·{\left(\ln v\right)}^{\frac{1}{2}}=\int \frac{1}{v}·{\left(\ln v\right)}^{\frac{1}{2}}dv$

$\Rightarrow u·{\left(\ln v\right)}^{\frac{1}{2}}=\frac{2}{3}{\left(\ln v\right)}^{\frac{3}{2}}+c .......\left(\mathrm{i}\right)$

Now using given value, $y=e^4-2\Rightarrow x=1$

$\therefore v=e^4 \Rightarrow u=5$

$5·\left(4^{\frac{1}{2}}\right)=\frac{2}{3}·{\left(4\right)}^{\frac{3}{2}}+c$

$\Rightarrow 10=\frac{16}{3}+c$

$\Rightarrow c=\frac{14}{3}$

Now finding, $y=e^9-2\Rightarrow v=y+2=e^9$

Now putting the value in equation $\left(\mathrm{i}\right)$ we get,

$\Rightarrow \mathrm{u}·3=\frac{2}{3}\times 27+\frac{14}{3}=18+\frac{14}{3}$

$\Rightarrow x+4=u=6+\frac{14}{9}$

$\Rightarrow x=2+\frac{14}{9}=\frac{32}{9}$