Question

Solve : $(1-x^2)\frac{dy}{dx}+2xy=x{(1-x^2)}^{1/2}$

Answer 1

Given,

$(1-x^2)\frac{dy}{dx}+2xy=x(1-x^2{)}^{\frac{1}{2}}$

$\frac{dy}{dx}+\frac{2x}{1-x^2}y=\frac{x}{\sqrt{1-x^2}}$

$\frac{dy}{dx}+Py=Q$

$P=\frac{2x}{1-x^2}\to I.F=e^{\int Pdx}=e^{\int \frac{2x}{1-x^2}dx}=e^{-\log (1-x^2)}=\frac{1}{1-x^2}$

$Q=\frac{x}{\sqrt{1-x^2}}$

$y\times I.F=\int Q\times I.Fdx$

$y\times \left(\frac{1}{1-x^2}\right)=\int \frac{x}{\sqrt{1-x^2}}\times \frac{1}{1-x^2}dx$

$y\left(\frac{1}{1-x^2}\right)=\int \frac{x}{(1-x^2{)}^{\frac{3}{2}}}dx$

$y\left(\frac{1}{1-x^2}\right)=\frac{1}{\sqrt{1-x^2}}+c$

$y=\sqrt{1-x^2}+C$