How to Solve this Second Order Differential Equation

I need to solve the following differential equation: $t^2x''(t)+tx'(t)-9x(t)=0$.

I am supposed to give solutions in the form $x(t)=at^b$, and I have no idea where to start, since the coefficients are variables instead of numbers. Any guidance would be greatly appreciated!

Sometimes the way to go is to just do it, and ask questions later. Plug in $y = at^b$ into the equation

$$t^2 \cdot b(b-1)at^{b-2} + t\cdot bat^{b-1} - 9at^b = 0$$

You'll see that the variable coefficients combine nicely with the derivatives. Divide through by $at^b$ to get $$b(b-1) + b - 9 = 0$$

Then solve for $b$. You end up with two linearly independent to solution for two values of $b$

Note: This a Cauchy-Euler equation of second order. Solutions to this type of equation are generally in the form of $at^b$ (with the exception of a multiple root, then you gain additional factors of $\ln t$)

We can use change of variables to solve it:

Set $u=\ln t, \Phi(u)=x(t)$ hence $x'(t)=(\Phi(u))'=\frac1t\Phi'(u)$ and $x''(t)=(\frac1t\Phi'(u))'=\frac1{t^2}(\Phi''(u)-\Phi'(u))$

Let's go back to our original equation and see what happens when we change it from $x$ to $t$ into $\Phi$ to $u$:

$$t^2x''(t)+tx'(t)-9x(9)=0\\t^2(\frac1{t^2}(\Phi''(u)-\Phi' (u)))+t(\frac1t\Phi'(u))-9\Phi(u)=0\\\Phi''(u)-\Phi' (u)+\Phi'(u)-9\Phi(u)=0\\\Phi''(u)-9\Phi(u)=0$$solving this we get $$\Phi(u)=c_1e^{3u}+c_2e^{-3u}$$change the variable to $t$ again to get $$x(t)=c_1e^{3\ln(t)}+c_2e^{-3\ln(t)}\\\boxed{=c_1t^{3}+c_2t^{-3}}$$