$\begin{aligned} & \frac{d y}{d x}+\frac{y}{1+x^2}=\frac{e^{\tan ^{-1} x}}{1+x^2} \\ & \text { I.F. }=e^{\int \frac{1}{1+x^2} d x}=e^{\tan ^{-1} x} \\ & y \cdot e^{\tan ^{-1} x}=\int\left(\frac{e^{\tan ^{-1} x}}{1+x^2}\right) e^{\tan ^{-1} x} \cdot d x \\ & \text { Let } \tan ^{-1} x=z \quad \therefore \frac{d x}{1+x^2}=d z \\ & \therefore y \cdot e^z=\int e^{2 z} d z=\frac{e^{2 z}}{2}+C \\ & y \cdot e^{\tan ^{-1} x}=\frac{e^{2 \tan ^{-1} x}}{2}+C \\ & \Rightarrow y=\frac{e^{\tan ^{-1} x}}{2}+\frac{C}{e^{\tan ^{-1} x}}\end{aligned}$
$\begin{aligned} & \because \mathrm{y}(1)=0 \Rightarrow 0=\frac{\mathrm{e}^{\pi / 4}}{2}+\frac{\mathrm{C}}{\mathrm{e}^{\pi / 4}} \Rightarrow \mathrm{C}=\frac{-\mathrm{e}^{\pi / 2}}{2} \\ & \therefore \mathrm{y}=\frac{\mathrm{e}^{\tan ^{-1} \mathrm{x}}}{2}-\frac{\mathrm{e}^{\pi / 2}}{2 \mathrm{e}^{\tan ^{-1} \mathrm{x}}} \\ & \therefore \mathrm{y}(0)=\frac{1-\mathrm{e}^{\pi / 2}}{2}\end{aligned}$