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

$y(1-x^2)=\large\frac{4x^3}{3}$=>$3y(1-x^2)=4x^3$

=>$y=\large\frac{4x^3}{3(1-x^2)}$