Ordinary Differential Equatiom

Question

$$\left(1-x^2\right) y''-4 x y'-\left(1+x^2\right) y=x $$

I am required to solve the above differential equation. Can't get around how to approach. Any help would be appreciated. $y' = \frac{dy}{dx}$

Answer 1

Hint

Try $$y(x)=\frac {z(x)}{1-x^2}$$ This will lead to a very simple differential equation in $z(x)$.

Answer 2

The method is the same as Claude Leibovici suggested, but maybe you should rather try the the following

$$z(x)=(1-x^2)y(x)$$ $$z'(x)=-2xy(x)+(1-x^2)y'(x)$$ $$z''(x)=-2y(x)-2xy'(x)-2xy'(x)+(1-x^2)y''(x)$$ $$z''(x)=(1-x^2)y''(x)-4xy'(x)-2y(x)$$

Adding $$y(x)(1-x^2)$$ on both sides results in

$$z''(x)+y(x)(1-x^2)=(1-x^2)y''(x)-4xy'(x)-(1+x^2)y(x)$$

Note that $y(x)(1-x^2)=z(x)$ and the right hand side is equal to $x$. Hence, you get the simple equation.

$$z''(x)+z(x)=x$$

Can you continue from here?