JEE Main 2024 — Differential Equations Question with Solution

Given: $\left(1+y^2\right)\left(1+{\log }_{e}x\right)dx+xdy=0$

$\Rightarrow \frac{\left(1+{\log }_{e}x\right)}{x}dx=\frac{-1}{\left(1+y^2\right)}dy$

$\Rightarrow \int \left(\frac{1}{x}+\frac{\log x}{x}\right)dx+\int \frac{dy}{1+y^2}=0$

$\Rightarrow \log x+\frac{(\log x{)}^2}{2}+{\tan }^{-1}y=\mathrm{C}$

This curve passes through the point $\left(1,1\right)$.

$\Rightarrow \log \left(1\right)+\frac{(\log \left(1\right){)}^2}{2}+{\tan }^{-1}\left(1\right)=C$

$\Rightarrow C=\frac{\pi }{4}$

$\Rightarrow \log x+\frac{(\log x{)}^2}{2}+{\tan }^{-1}y=\frac{\pi }{4}$

Now, putting $x=e$

$\Rightarrow \log e+\frac{(\log e{)}^2}{2}+{\tan }^{-1}y=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}y=\frac{\pi }{4}-\frac{3}{2}$

$\Rightarrow y=\tan \left(\frac{\pi }{4}-\frac{3}{2}\right)$

$\Rightarrow y=\frac{\tan \frac{\pi }{4}-\tan \left(\frac{3}{2}\right)}{1+\tan \frac{\pi }{4}\tan \left(\frac{3}{2}\right)}$

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

Hence, on comparing we get,

$\Rightarrow \alpha =\beta =1$

$\Rightarrow \alpha +2\beta =3$