Question

Determine the general solution of y' + [1-3x^2/x(1-x^2)]y = 1/1-x^2

Answer 1

Given

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

This is a linear equation with

$p(x)=\frac{1-3x^2}{x(1-x^2)},\ \ \ q(x) =\frac{1} {1-x^2}$

Since

$\begin{aligned} \frac{1-3x^2}{x(1-x^2)}&=\frac{A}{x }+\frac{B}{1-x}+\frac{C}{1+x}\\ &= \frac{A (1-x^2)+Bx(1+x)+Cx(1-x)}{x(1-x^2)}\\ 1-3x^2&=A (1-x^2)+Bx(1+x)+Cx(1-x) \end{aligned}$

Then

$\begin{aligned} \text{At } \ x&=0\ \ \ \to \ A=1\\ \text{At } \ x&=-1\ \ \ \to \ C=1\\ \text{At } \ x&= 1\ \ \ \to \ B=-1 \end{aligned}$

Hence

$\begin{aligned} \int p(x)dx&=\int \frac{1-3x^2}{x(1-x^2)}dx\\ &=\int \frac{dx}{x }-\int \frac{dx}{1-x}+\int \frac{dx}{1+x}\\ &=\ln x+\ln (1-x)+\ln (1+x)\\ &=\ln x(1-x^2) \end{aligned}$

Since the integral factor given by

$\begin{aligned} \mu&=e^{\int p(x)dx}\\ &=e^{\ln x(1-x^2)}\\ &= x(1-x^2) \end{aligned}$

Then the solution is

$\begin{aligned} \mu y&=\int \mu q(x)dx\\ x(1-x^2) y &= \int x(1-x^2)\frac{1}{1-x^2}dx\\ &=\int xdx\\ &= \frac{1}{2}x^2+C \end{aligned}$

Hence

$y= \frac{ x^2}{2x(1-x^2)}+\frac{ C }{2x(1-x^2)}$