Answer
\[y(x)=a(1+\sqrt{1-x^2})\]
Work Step by Step
$(1-x^2)y'+xy=ax$ _____(1)
$(1-x^2)y'=x(a-y)$
$(1-x^2)\frac{dy}{dx}=x(a-y)$
Separating variables,
$\frac{x}{1-x^2}dx=\frac{dy}{a-y}$
Integrating,
$C+\int\frac{x}{1-x^2}dx=\int\frac{dy}{a-y}$
$C$ is constant of integration
$C-\frac{1}{2}\int\frac{-2x}{1-x^2}dx=-\ln|a-y|$
$C-\frac{1}{2}\ln|1-x^2|=\ln\left|\frac{1}{a-y}\right|$
$\ln\left|\frac{\sqrt{1-x^2}}{a-y}\right|=C$
Using initial condition $y(0)=2a$
$\ln\left|\frac{1}{a-2a}\right|=C$
$\ln\left|\frac{-1}{a}\right|=C$ ___(3)
Using (2) and (3)
$\frac{\sqrt{1-x^2}}{a-y}=\frac{-1}{a}$
$\Rightarrow a\sqrt{1-x^2}=y-a$
$\Rightarrow a(1+\sqrt{1-x^2})=y(x)$
Hence solution fo given initial value problem is $y(x)=a(1+\sqrt{1-x^2})$
