General Solution of Linear Homogenous Differential Equation The function $y_1=x^\frac{1}{2}$ is a solution of $4x^2y''+y=0$. What is the general solution of the linear homogenous differential equation on the interval $(0,\infty)$.

Can someone help me out. I am preparing for my test and don't know how to do this.
We have 
\begin{align}
4x^2y''\left(x\right)+y\left(x\right)=0.\tag{1}
\end{align}
This is a classic example of a Cauchy-Euler equidimensional equation. Substitute $y\left(x\right)=x^r$. Then
\begin{align}
y\left(x\right)=x^r,\\
y'\left(x\right)=rx^{r}x^{-1},\\
y''\left(x\right)=r\left(r-1\right)x^{r}x^{-2}.
\end{align}
This gives us
\begin{align}
0&amp;=4x^2\left(r\left(r-1\right)x^{r}x^{-2}\right)+\left(x^{r}\right)\\
&amp;=4r\left(r-1\right)x^r+x^r\\
&amp;=4r\left(r-1\right)+1.\tag{2}
\end{align}
Now solve this characteristic equation for $r$:
\begin{align}
r&amp;=\frac{4\pm \sqrt{16-4\left(4\right)\left(1\right)}}{8}\\
&amp;=\frac{4\pm 0}{8}=\frac{1}{2}.\tag{3}
\end{align}
Therefore we have $y\left(x\right)=C_0 x^{1/2}$ as a general solution, but one of repeated roots. To find the second solution you make the observation that $\frac{\partial }{\partial r}\left(x^r\right)=x^r\log\left(x\right)$. Therefore, the most "general" solution is 
\begin{align}
y\left(x\right)=C_0x^{1/2}+C_1x^{1/2}\log\left(x\right),\;\;x\gt 0.\tag{4}
\end{align}
Hint: You can use Frobenius Method.