Solution of differential equation $\frac{dy}{dx}+\frac{x}{1-x^2}y=x\sqrt{y}$ is:

$3\sqrt{y}=-(1-x^2)+c(1-x^2)^{1/4}$

$\frac{dy}{dx}+\frac{x}{1-x^2}y=x\sqrt{y}$ [Linear D.E.] $\frac{1}{\sqrt{y}}\frac{dy}{dx}+(\frac{x}{1-x^2})\sqrt{y}=x$

Substitute $\sqrt{y}=t;\frac{1}{2\sqrt{y}}\frac{dy}{dx}=\frac{dt}{dx}⇒\frac{dt}{dx}+(\frac{x}{1-x^2})\frac{t}{2}=\frac{x}{2}$

$I.F. =e^{\int\frac{x}{2(1-x^2)}}dx=e^{-\frac{1}{4}ln(1-x^2)}=\frac{1}{(1-x^2)^{1/4}}$

Hence the general solution is: $\sqrt{y}.\frac{1}{(1-x^2)^{1/4}}=\int\frac{x}{2(1-x^2)^{1/4}}dx+c$

$⇒\frac{\sqrt{y}}{(1-x^2)^{1/4}}=-\frac{1}{3}(1-x^2)^{3/4}+c⇒3\sqrt{y}=-(1-x^2)+c(1-x^2)^{1/4}$