Mathematics - Differential Equation Question with Solution | TestHub

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