JEE Main 2024 — 6 April, Shift 1 — Question 17

$\frac{d y}{d x}+\frac{y}{1+x^{2}}=\frac{e^{\tan ^{-1} x}}{1+x^{2}}$

I.F. $=\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}}=\mathrm{e}^{\tan ^{-1} \mathrm{x}}$

$y \cdot e^{\tan ^{-1} x}=\int\left(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\right) e^{\tan ^{-1} x} \cdot d x$

Let $\tan ^{-1} \mathrm{x}=\mathrm{z} \quad \therefore \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}=\mathrm{dz}$

$\therefore y . e^{z}=\int e^{2 z} d z=\frac{e^{2 z}}{2}+C$

$y \cdot e^{\tan ^{-1} x}=\frac{e^{2 \tan ^{-1} x}}{2}+C$

$\Rightarrow y=\frac{e^{\tan ^{-1} x}}{2}+\frac{C}{e^{\tan ^{-1} x}}$

$\because y(1)=0 \Rightarrow 0=\frac{\mathrm{e}^{\pi / 4}}{2}+\frac{\mathrm{C}}{\mathrm{e}^{\pi / 4}} \Rightarrow \mathrm{C}=\frac{-\mathrm{e}^{\pi / 2}}{2}$

$\therefore y=\frac{e^{\tan ^{-1} x}}{2}-\frac{e^{\pi / 2}}{2 e^{\tan ^{-1} x}}$

$\therefore \mathrm{y}(0)=\frac{1-\mathrm{e}^{\pi / 2}}{2}$