Solving $(x^2-1)\ddot y-2x\dot y +2y=1$

Question

Solving $$(x^2-1)\ddot y-2x\dot y +2y=1$$

I've solved the homogenous equation:

$$y=A(x^2+1)+Bx$$

Where A and B are constants of integration, but I can't for the life of me seem to remember how to find the private solution when coefficients aren't constants. I'd appreciate any hints.

Ron

Answer 1

I presume you mean $(x^2- 1)y''- 2xy'- y= 1$. An obvious thing to try, since the right hand side is a constant, is y= constant. With y= C, y'= y''= 0 so the equation becomes -C= 1 or C= -1. y(x)= -1 satisfies this equation.

Answer 2

By setting $y=u+\frac{1}{2}$ we have $$(x^2-1)u''-2xu' +1(1+1)u=0$$ which is a Legendre ODE.

Answer 3

$$(x^2-1)\ddot y-2x\dot y +2y=1$$ An obvious particular solution is $y=\frac{1}{2}$

Add it to the solution of the homogeneous ODE : $$y=A(x^2+1)+Bx+\frac{1}{2}$$