The general solution of the differential equation \(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{\frac{1}{2}}\) is

\(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{\frac{1}{2}} \)
\( \therefore \frac{d y}{d x}+\frac{2 x y}{1-x^{2}}=\frac{x}{\left(1-x^{2}\right)^{\frac{1}{2}}} \)
\( \text { I.F. }= e^{\int \frac{2 x}{1-x^{2}} d x}=e^{-\int \frac{-2 x}{1-x^{2}} d x}=e^{-\log \left(1-x^{2}\right)}=\) \(e^{\log \left(\frac{1}{1-x^{2}}\right)}=\frac{1}{1-x^{2}}\)
\(\therefore\left(\frac{1}{1-x^{2}}\right) =\int \frac{x}{\left(1-x^{2}\right)^{\frac{1}{2}}} \times \frac{1}{\left(1-x^{2}\right)} d x \)
\( =\frac{-1}{2} \int \frac{-2 x}{\left(1-x^{2}\right)^{3 / 2}} d x=\left(-\frac{1}{2}\right) \frac{\left(1-x^{2}\right)^{-\frac{1}{2}}}{\left(-\frac{1}{2}\right)}+c \)
\( y\left(\frac{1}{1-x^{2}}\right) =\frac{1}{\sqrt{1-x^{2}}+c \Rightarrow y}=\sqrt{1-x^{2}}+c\left(1-x^{2}\right)\)