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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!

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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 $\ln t$)

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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 $$x(t)=c_1e^{3\ln(t)}+c_2e^{-3\ln(t)}\\\boxed{=c_1t^{3}+c_2t^{-3}}$$

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