Correct option is (2) \(\frac{1}{2}\left(1-e^{\pi / 2}\right)\)
\(\frac{d y}{d x}+\frac{y}{1+x^{2}}=\frac{e^{\tan ^{-1} x}}{1+x^{2}}\)
\( I.F. =\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}}=\mathrm{e}^{\tan ^{-1} \mathrm{x}}\)
\(y \cdot e^{\tan ^{-1} x}=\int\left(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\right) e^{\tan ^{-1} x} \cdot d x\)
\( Let \tan ^{-1} \mathrm{x}=\mathrm{z} \quad \therefore \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}=\mathrm{dz}\)
\(\therefore y \cdot \mathrm{e}^{\mathrm{z}}=\int \mathrm{e}^{2 \mathrm{z}} \mathrm{dz}=\frac{\mathrm{e}^{2 \mathrm{z}}}{2}+\mathrm{C}\)
\(y \cdot \mathrm{e}^{\tan ^{-1} \mathrm{x}}=\frac{\mathrm{e}^{2 \tan ^{-1} \mathrm{x}}}{2}+\mathrm{C}\)
\(\Rightarrow y=\frac{\mathrm{e}^{\tan ^{-1} x}}{2}+\frac{\mathrm{C}}{\mathrm{e}^{\tan ^{-1} \mathrm{x}}}\)
\( \because y(1)=0 \Rightarrow 0=\frac{\mathrm{e}^{\pi / 4}}{2}+\frac{\mathrm{C}}{\mathrm{e}^{\pi / 4}} \Rightarrow \mathrm{C}=\frac{-\mathrm{e}^{\pi / 2}}{2}\)
\(\therefore y=\frac{e^{\tan ^{-1} x}}{2}-\frac{e^{\pi / 2}}{2 e^{\tan ^{-1} x}}\)
\(\therefore \mathrm{y}(0)=\frac{1-\mathrm{e}^{\pi / 2}}{2}\)
