JEE Main 2025 — Differential Equations Question with Solution

$\begin{aligned} & y^2 d x+\left(x-\frac{1}{y}\right) d y=0 \\ & y^2 d x=\left(\frac{1}{y}-x\right) d y \\ & \Rightarrow y^2 \frac{d x}{d y}=\frac{1}{y}-x \\ & \Rightarrow \frac{d x}{d y}+\frac{x}{y^2}=\frac{1}{y^3} \\ & \text { I.F. }=e^{\int \frac{1}{y^2} d y}=e^{\frac{-1}{y}} \end{aligned}$ $\therefore$ Solution is $x{e}^{\frac{-1}{y}}=\int e^{-\frac{1}{y}}\times \frac{1}{y^3}dy+C$ Let $\frac{-1}{y}=t$ $\begin{aligned} & \Rightarrow \frac{1}{y^2} d y=d t \\ & \Rightarrow x e^{-\frac{1}{y}}=-\int e^t t d t+C \\ & \Rightarrow x e^{-\frac{1}{y}}=-e^t(t-1)+C \\ & \Rightarrow x e^{-\frac{1}{y}}=-e^{\frac{-1}{y}}\left(\frac{-1}{y}-1\right)+C \end{aligned}$ $x(1)=1$ $\Rightarrow e^{-1}=-e^{-1}(-2)+C$ $\Rightarrow C=-e^{-1}$ $\Rightarrow x=\frac{1}{y}+1-e^{-1+\frac{1}{y}}$ $x\left(\frac{1}{2}\right)=3-e$