Solving differential equation (t^2-1)y''-6y=1

[CGandC]

Homework Statement

Solve the $\mathrm{DE}\left(t^2-1\right) \ddot{y}-6 y=1$ if it is known that the corresponding homogeneous problem has a polynomial particular solution.

Relevant Equations

I haven't learned about frobenius method.
I've learned about variation of parameters and I think that's the key here.

This was a question from some past exam, I found online a solution but it uses Frobenius method which wasn't taught in the course.

I would approach the solution by attempting to find a solution to the homogeneuous DE $\left(t^2-1\right) \ddot{y}-6 y=0$, but that by itself is quite tricky. Using the hint, I've found a homogeneuous solution $y_{homogeneuous_1} = t- t^3$, and a particular solution $y_{particular} = -1/6$ to the non-homogeneuous equation.



However, I'm unable to find a general solution which is of the form $y = c_1\cdot y_{homogeneuous_1} + c_2 \cdot y_{homogeneuous_2} + y_{particular}$ since I lack another homogeneuous solution $y_{homogeneuous_2}$ ( which should be linearly independent to $y_{homogeneuous_1}$ ) which I can't find; I tried approaching the solution to the last problem using the fact ( variation of parameters ) that $y_{particular} = a_1(x)\cdot y_{homogeneuous_1} + a_2(x) \cdot y_{homogeneuous_2}$, but I got stuck.



Thanks in advance for the help!

 

[pasmith]

If $y_1$ is a solution of the homogenous equation $$y'' + py' + q = 0$$ then $$y_2 = y_1 \int \frac{W}{y_1^2}\,dx$$ is a linearly independent solution, where the Wronskian $W$ satisfies $W' + pW = 0$. Since here $p = 0$ you can take $W = 1$.

 

[CGandC]

Thank you sir! I understand now