Step 1:
Given equation of the curve $(1-x^2)\large\frac{dy}{dx}$$-2xy=4x^2$
Divide throughout by $(1-x^2)$
$\large\frac{dy}{dx}-\frac{2x}{1-x^2}.$$y=\large\frac{4x^2}{1-x^2}$
Clearly this is a linear differential equation of the form
$\large\frac{dy}{dx}$$+P.y=Q$
Where $P=-\large\frac{2x}{1-x^2}$ and $ Q=\large\frac{4x^2}{1-x^2}$
Let us now find the integrating factor I.F
$e^{\int pdx}=e^{\int \Large\frac{-2x}{1-x^2}}dx$
Step 2:
Consider $\int \large\frac{-2x}{1-x^2}$$dx$
Put $ 1-x^2=t$ on differentiating w.r.t t
We get $-2xdx=dt$
Now substituting this we get,
$\large\int \frac{dt}{t}$$=\log |t|$
Substituting for t we get,
$\log |1-x^2|$
Hence I.F is $e^{\large\log(1-x^2)}$
but $e^{\large\log x}$$=x$
Hence $I.F=1-x^2$
Now the required solution is
$ye^{\large\int pdx}$$=\int Q. e ^{\large\int pdx}$$dx+c$
$y.(1-x^2)=\int \large\frac{4x^2}{1-x^2}$$(1-x^2)dx+c$
=>$y(1-x^2)=\int 4x^2.dx+c$
On integrating we get
$y(1-x^2)=\large\frac{4x^3}{3}+c$
Step 3:
It is given that the curve passes through the orgin, which means $x=0,y=0$
Let us now substitute for x and y to evaluate the value for c
$0(1+0)=4 \times 0+c=>c=0$
Hence the required solution is
$y(1-x^2)=\large\frac{4x^3}{3}$=>$3y(1-x^2)=4x^3$
=>$y=\large\frac{4x^3}{3(1-x^2)}$