shamieh said:
The D.E. $t^2y'' -2y = 0$ has two solutions in the form $y = t^p$. Find them.
I'm not sure how to do this one. Shouldn't they give me a $y_1(t)$ = something and a $y_2(t)$ = to something? Am I given enough information?
Well if $\displaystyle \begin{align*} y = t^p \end{align*}$ then $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} = p\,t^{p-1} \end{align*}$ and $\displaystyle \begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = p \, \left( p - 1 \right) \, t^{p-2} \end{align*}$. Substituting into the DE gives
$\displaystyle \begin{align*} t^2 \, p \, \left( p - 1 \right) \, t^{p-2} - 2\,t^p &= 0 \\ p\,\left( p - 1 \right) \, t^p - 2\,t^p &= 0 \\ t^p \, \left[ p\, \left( p - 1 \right) - 2 \right] &= 0 \\ t^p \, \left( p^2 - p - 2 \right) &= 0 \\ t^p \, \left( p - 2 \right) \left( p + 1 \right) &= 0 \end{align*}$
So that means either $\displaystyle \begin{align*} t^p = 0 \implies t = 0 \end{align*}$ is a solution, or $\displaystyle \begin{align*} p - 2 = 0 \implies p = 2 \end{align*}$ and $\displaystyle \begin{align*} p + 1 = 0 \implies p = -1 \end{align*}$ also give solutions to the DE.
