4zy''+2y'+y=0
find series solution about z=0 and identify these solution in closed form
If you divide through by x you get $\displaystyle 4y'' + \frac{2}{x}y'+\frac{1}{x}y = 0$. The coefficient function for $\displaystyle y'$ is Laurent series of order $\displaystyle 1$ and same for $\displaystyle y$. This means the equation is solvable through Frobenius method. The Frobenius equation would be $\displaystyle 4k(k-1)+2k = 0$ so $\displaystyle k=0,2/3$. These do not differ by integers so you can solve this equation on $\displaystyle (0,\infty)$ by letting $\displaystyle y = x^{k}\sum_{n=0}^{\infty} a_n x^n$ where $\displaystyle k=0,2/3$. For each value you will get a new linearly independent solution.
Hello,
Let $\displaystyle y(z)=\sum_{n \ge 0} a_n z^n$
You will have to determine $\displaystyle a_n$ ~
$\displaystyle y'(z)=\sum_{n \ge 1} n a_n z^{n-1}$
$\displaystyle y''(z)=\sum_{n \ge 2} n(n-1) a_n z^{n-2}$
$\displaystyle \implies 4zy''(z)+2y'(z)+y(z)=0$
$\displaystyle 4 \sum_{n \ge 2} n(n-1) a_n z^{n-1}+2 \sum_{n \ge 1} n a_n z^{n-1}+\sum_{n \ge 0} a_n z^n=0$
Putting all the powers to $\displaystyle z^n$, by changing the indices ~
In the first and second ones, n-1 becomes n. Then the series will start at 1 and 0..
$\displaystyle 4 \sum_{n \ge 1} n(n+1) a_{n+1} z^n+2 \sum_{n \ge 0} (n+1) a_{n +1} z^n+\sum_{n \ge 0} a_n z^n=0$
Now, starting the series at the same point :
$\displaystyle 4 \sum_{n \ge 1} n(n+1) a_{n+1} z^n+2 \sum_{n \ge {\color{red}1}} (n+1) a_{n+1} z^n+{\color{red}a_1}+\sum_{n \ge {\color{red}1}} a_n z^n+{\color{red}a_0}=0$
Therefore :
$\displaystyle a_0+a_1+\sum_{n \ge 1} \left[4n(n+1)a_{n+1}+2(n+1)a_{n+1}+a_n\right] z^n=0$
--> $\displaystyle a_0+a_1=0$
And :
$\displaystyle 4n(n+1)a_{n+1}+2(n+1)a_{n+1}+a_n=0$
Find the induction formula