JEE Main 2026 Jan 24 Shift 2, Mathematics Q16: Methods of Solving First Order, First Degree Differential Equation

The limit is $\frac{0}{0}$ form as $t\to x$. Apply L'Hopital with respect to $t$:

$\underset{t\to x}{\lim }\frac{2ty(x)-x^2{y}^{\prime }(t)}{-1}=3\Rightarrow -(2xy(x)-x^2{y}^{\prime }(x))=3\Rightarrow x^2{y}^{\prime }(x)-2xy(x)=3$.

This is linear: $y^{\prime }(x)-\frac{2}{x}y(x)=\frac{3}{x^2}$. Integrating factor $=e^{-\int \frac{2}{x}dx}=\frac{1}{x^2}$.

$\frac{d}{dx}\left(\frac{y}{x^2}\right)=\frac{3}{x^4}\Rightarrow \frac{y}{x^2}=-\frac{1}{x^3}+C\Rightarrow y=C{x}^2-\frac{1}{x}$.

Use $y(1)=2$: $C-1=2\Rightarrow C=3$. So $y(x)=3x^2-\frac{1}{x}$.

$y(2)=12-\frac{1}{2}=\frac{23}{2}$. Therefore $2y(2)=23$.