JEE 2025 — Mathematics PYQ

$$y^{\prime }dx+\left(x-\frac{1}{y}\right)dy=0$$
$$y^{\prime }dx+x\frac{dy}{y}=1$$
$$\frac{dx}{dy}\left(\frac{1}{x}\right)=x-\frac{1}{y}$$
$$P=\frac{1}{y^2},Q=\frac{1}{y^2}$$
$$1.F=e^{\int \frac{1}{y^2}dy}=e^{\frac{1}{y^2}}=e^{-\frac{1}{y}}$$
$$\text{Solution of DE}$$
$$x(1-F)=\int Q\cdot (1-F)dy+C$$
$$x{e}^{-\frac{1}{y}}=\int -\frac{1}{y}{e}^{-\frac{1}{y}}dy+C$$
$$e^{-\frac{1}{y}}=t\Rightarrow -\frac{1}{y}dt=dt$$
$$e^{-\frac{1}{y}}=\int -\ln tdt+C$$
$$x{e}^{-\frac{1}{y}}=-t(\ln t-1)+C$$
$$x{e}^{-\frac{1}{y}}=-e^{-\frac{1}{y}}\left(\frac{1}{y}-1\right)+C$$
$$x=\frac{1}{y}+1+C{e}^{\frac{1}{y}}$$
$$x(1)=1=1+1+Ce\Rightarrow C=\frac{-1}{e}$$
$$\therefore x\left(\frac{1}{e}\right)=2+1+\left(\frac{1}{e}\right)e^2=3-e$$