Question

Let $y=y\left(x\right)$ be the solution of the differential equation $x(1-x^2)\frac{dy}{dx}+(3x^{2}y-y-4x^3)=0,x>1$ with $y\left(2\right)=-2$. Then $y\left(3\right)$ is equal to

Answer 1

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

$IF=e^{\int \frac{3x^2-1}{x-x^3}dx}=e^{-\ln |x^3-x|}=e^{-\ln (x^3-x)}$

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

Solution of D.E. can be given by

$y\cdot \frac{1}{x^3-x}=\int \frac{4x^3}{x(1-x^2)}\cdot \frac{1}{x(x^2-1)}dx$

$\Rightarrow \frac{y}{x^3-x}=\int \frac{-4x}{{(x^2-1)}^2}dx$

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

at $x=2,y=-2$

$\frac{-2}{6}=\frac{2}{3}+c\Rightarrow c=-1$

$\text{at}x=3\Rightarrow \frac{y}{24}=\frac{2}{8}-1\Rightarrow y=-18$