Solving 2nd Order non homogeneous differential equation using Wronskian when one solution is given

Question

> Solve by variation of parameters $x^2 y'' + 2xy' - 6y = 5x^4$ and given that $x^2$ is a solution of homogeneous equation.

Hint: Find other fundamental solution using Wronsian.

My Attempt

Answer 1

Given $y_1=x^2$ is a solution, you can reduce the order by substituting $y(x)=x^2v(x)$ to make it easier to find the other fundamental solution.

$$y=x^2v \implies y' = 2xv + x^2v' \implies y'' = 2v + 4xv'+ x^2v''$$

$$\begin{align*} 0 &= x^2y'' + 2xy' - 6y \\[1ex] &= x^2(2v+4xv'+x^2v'') + 2x(2xv+x^2v') -6x^2v \\[1ex] &= 6x^3v'+x^4v'' \\[1ex] &= xw' + 6w & w(x)=v'(x) \end{align*}$$

Solve for $w\to v\to y$.

$$\begin{align*} w(x) &= Cx^{-6} \\ v(x) &= C_1x^{-5} + C_2 \\ y(x) &= C_1x^{-3} + C_2x^2 \end{align*}$$

The solution $y_1=x^2$ is already accounted for, so the other is $y_2=x^{-3}$.

Now the Wronskian is

$$W(y_1,y_2) = \begin{vmatrix} y_1 & y_2 \\ {y_1}' & {y_2}' \end{vmatrix} = -5x^{-2}$$

and you can proceed with variation of parameters.