If \(\frac{d y}{d x}+\left(\frac{x}{1+x^2}\right) y=\frac{\sqrt{x}}{\sqrt{1+x^2}} ; y(0)=0\), then \(y(1)\) will be (24 …

\(\begin{aligned}
& \frac{d y}{d x}+\left(\frac{x}{1+x^2}\right) y=\frac{\sqrt{x}}{\sqrt{1+x^2}}; \quad y(0)=0 \\
& \text { I.f} =e^{\int \frac{x}{1+x^2} d x} e^{\frac{1}{2} \ln \left(1+x^2\right)} \\
& I.f=\sqrt{1+x^2} \\
& y \sqrt{1+x^2}=\int \sqrt{x} d x \\
& y \sqrt{1+x^2}=\frac{2}{3} x^{\frac{3}{2}}+C \\
& x=0, \quad y=0 \\
& c=0 \\
& y \sqrt{1+x^2}=\frac{2}{3} x^{\frac{3}{2}} \\
& x=1 \\
& y \cdot \sqrt{2}=\frac{2}{3} \\
& y=\frac{2}{3 \sqrt{2}}
\end{aligned}\)