Solve the equation. Use the method of Problem 34 to solve the given equation for t > 0. No need to do an actual change of variables, just use the result the Problem 34.
\begin{align*}
\ t^2 y''+ty' + y = 0,\\
\end{align*}
Solution:
$\text{According to Problem 34, if we have an Euler equation in the form of } t^2 y''+aty' + by = 0\\
\text{Then by changing variables } x = \ln(t) \text{, we can transform the equation into } \frac{d^2y}{dx^2} + (a-1)\frac{dy}{dx} + by = 0$
\begin{align*}
\text{In our case, } t^2 y''+ty' + y = 0 \text{ is an Euler eqaution where a = 1, b = 1 } \\
\text{Then by changing variables } x = \ln(t), \\
\text{we can transform the equation into } y(x)'' + (1-1)y(x)' + y = 0\\
Then \ y(x)'' + y = 0 \\
\text{To solve }y(x)'' + y = 0, \text{assume } y =e^{rx} \text{ is a solution,}\\
\text{and we will have } y'=re^{rx}, y''&=r^2e^{rx}\\
Then \ y(x)'' + y = 0 &\Rightarrow r^2e^{rx}+re^{rx}=0\\
e^{rx}(r^2+1) &= 0\\
\text{Since }e^{rx} \neq 0, \text{then } r^2+1 = 0\\
r^2 &= \sqrt{-1}\\
r &= \pm i \text{ where }\lambda =0 \ and \ \mu =1 \\
y(x) &= c_{1} e^{\lambda x} cos(\mu x) + c_{2} e^{\lambda x} sin(\mu x) \\
&= c_{1} e^{0} cos(x) + c_{2} e^{0} sin(x)\\
&= c_{1}cos(x) + c_{2}sin(x) \\
\text{Since we let }x = \ln(t) \text{ at the beginning, }\\
\text{Then we will have, } \\
y(t) &= c_{1}cos(\ln t) + c_{2}sin(\ln t) \\
&\text{as the general solution.} \\
\end{align*}
