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Q.
Let $y=y(x)$ be the solution of the differential equation $x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2} y-y-4 x^{3}\right)=0, x >1$ with $y(2)=-2$. Then $y(3)$ is equal to
$x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2} y-y-4 x^{3}\right)=0$
$x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2}-1\right) y=4 x^{3}$
$\frac{d y}{d x}+\frac{\left(3 x^{2}-1\right)}{\left(x-x^{3}\right)} y=\frac{4 x^{3}}{\left(x-x^{3}\right)}$
$\frac{d y}{d x}+P y=Q$
$I F=e^{\int P d x}=e^{\int \frac{3 x^{2}-1}{x-x^{3}} d x}$
$x-x^{3}=t \Rightarrow I F=e^{\int \frac{-d t}{t}}$
$=e^{-\text{lnt}}=\frac{1}{t}$
$\therefore I F=\frac{1}{x-x^{3}}$
$y \times I F=\int Q \times I F d x$
$y \left(\frac{1}{ x - x ^{3}}\right)=\int \frac{4 x ^{3}}{ x - x ^{3}} \times \frac{1}{\left( x - x ^{3}\right)} dx$
$=\int \frac{4 x ^{3}}{\left( x - x ^{3}\right)^{2}} dx$
$=\int \frac{4 x }{\left(1- x ^{2}\right)^{2}} dx\,], 1- x ^{2}= K$
$=2 \int \frac{- dK }{ K ^{2}}\,], -2 xdx = dK$
$=-2\left(-\frac{1}{ K }\right)+ c$
$\frac{ y }{ x - x ^{3}}=\frac{2}{ K }+ c$
$\frac{ y }{ x - x ^{3}}=\frac{2}{1- x ^{2}}+ c$
At $x =2, y =-2$
$\frac{-2}{2-8}=\frac{2}{1-4}+ c$
$\frac{1}{3}=\frac{-2}{3}+ c$
$\therefore C =1$
$\frac{ y }{ x - x ^{3}}=\frac{2}{1- x ^{2}}+1$
Put $x=3$
$\frac{y}{3-27}=\frac{2}{1-9}+1$
$\frac{y}{-24}=-\frac{1}{4}+1$
$\frac{y}{-24}=\frac{3}{4}$
$y=\frac{3}{4}(-24)=-18$