Answer
$y_1(t) = t^{-2}$ and $y_2(t) = t^{-2} \ln(t)$ are solutions to the differential equation
$t^{2}y'' + 5ty' +4y = 0$, $t>0$.
Work Step by Step
We must show that $y_{1}(t) = t^{-2}$ and $y_{2}(t) = t^{-2}\ln(t)$ are both solutions to the given differential equation, $t^{2}y'' + 5ty'' + 4y = 0$, $t>0$.
1) To show this, we first differentiate $y_{1}$ w.r.t. $t$ twice:
$y_{1}(t) = t^{-2}$
$y_{1}'(t) = -2t^{-3}$
$y_{1}''(t) = 6t^{-4}$.
Substituting $y_1$ for $y$ in the original equation:
$t^{2}(6t^{-4})+5t(-2t^{-3})+4t^{-2}$
$= 6t^{-2}-10t^{-2}+4t^{-2}$
$= 0$.
We have thus demonstrated that $y_{1}$ is a solution of our differential equation.
2) We next use the same procedure for $y_{2}$:
$y_{2}(t) = t^{-2}\ln(t)$.
$y_2'(t) = t^{-2}\cdot \frac{1}{t} + \ln t \cdot (-2t^{-3})$
$=t^{-3} - 2t^{-3}\ln t$.
$y_{2}''(t) = -3t^{-4} - (2t^{-3}\cdot t^{-1} - 6t^{-4}\ln t)$
$= -3t^{-4} - 2t^{-4} + 6t^{-4}\ln t$
$= -5t^{-4}+6t^{-4}\ln t$.
Substituting $y_{2}$ for $y$ in the given equation:
$t^{2}y_{2}''(t) + 5ty_{2}'(t) + 4y_{2}(t)$
$= t^{2}(-5t^{-4} + 6t^{-4}\ln t) +5t(t^{-3} - 2t^{-3}\ln t) + 4t^{-2}\ln t$
$= -5t^{-2} + 6t^{-2}\ln t + 5t^{-2} - 10t^{-2}\ln t +4t^{-2}\ln t$
$= 0$, which was to be shown.
We have thus demonstrated that both $y_{1}$ and $y_{2}$ are solutions to our given differential equation.
