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 …

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