Let y = y x be the solution of the differential equation x 1 - x 2 d y d x + 3 x 2 y - y - 4 x 3 = 0 , x…

Given $x\left(1-x^2\right)\frac{dy}{dx}+\left(3x^{2}y-y-4x^3\right)=0$

$\left(x-x^3\right)\frac{dy}{dx}+\left(3x^2-1\right)y=4x^3$

$\frac{dy}{dx}+\frac{\left(3x^2-1\right)}{\left(x-x^3\right)}y=\frac{4x^3}{\left(x-x^3\right)}$

This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$

Here $IF=e^{\int Pdx}=e^{\int \frac{3x^2-1}{x-x^3}dx}$

Let $x-x^3=t\Rightarrow IF={\mathrm{e}}^{\int -\frac{\mathrm{dt}}{\mathrm{t}}}$

$=e^{-\ln t}=\frac{1}{t}$

$\therefore IF=\frac{1}{x-x^3}$

So the general solution of the differential equation will be 

$y\times IF=\int Q\times IFdx$

$y\left(\frac{1}{x-x^3}\right)=\int \frac{4x^3}{x-x^3}\times \frac{1}{\left(x-x^3\right)}dx$

$=\int \frac{4x^3}{{\left(x-x^3\right)}^2}dx$

$=\int \frac{4x}{{\left(1-x^2\right)}^2}dx$

Now let $I=\int \frac{4x}{{\left(1-x^2\right)}^2}dx$

Substituting $1-x^2=z$

$I=2\int \frac{-dz}{z^2}$

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

Hence the solution of the differential equation becomes 

$\frac{y}{x-x^3}=\frac{2}{1-x^2}+c$

At $x=2, y=-2$

$\frac{-2}{2-8}=\frac{2}{1-4}+c$

$\frac{1}{3}=\frac{-2}{3}+c$

$\therefore c=1$

Hence the particular solution will be $\frac{y}{x-x^3}=\frac{2}{1-x^2}+1$

Put $x=3$

$\frac{y}{3-27}=\frac{2}{1-9}+1$

$\frac{y}{-24}=-\frac{1}{4}+1$

$-\frac{y}{24}=\frac{3}{4}$

$y=\frac{3}{4}\left(-24\right)=-18$