# Thread: power series solutions

1. ## power series solutions

$xy'=y$
I have to find a power series solution to this equation of the form $\sum a_{n}x^{n}$
Then solve the equation directly.

I have worked out by studying the equation that the solution is $nx$ but i can't work out how to obtain that answer.

My working so far is:
$xy'-y=0$
$y'-\frac{y}{x}=0$
$\sum na_{n}x^{n-1} - \frac{1}{x}\sum a_{n}x^{n}=0$
(both of sums go from $n=0$ to $\infty$ I wasn't sure how to write those in LaTex.)

2. I solve the wrong equation

3. Hello,

For this kind of problems, you should write the starting indice
It's important.

Let $y=\sum_{n\geq 0} a_nx^n$ be the solution of the equation.

$y=a_0+\sum_{n\geq 1}a_nx^n$. Then $y'=\sum_{n\geq 1}na_nx^{n-1}$

\begin{aligned}xy'=y &\Longrightarrow x\sum_{n\geq 1} na_nx^{n-1}=\sum_{n\geq 0} a_nx^n \\
&\Longrightarrow \sum_{n\geq 1} na_nx^n=a_0+\sum_{n\geq 1} a_nx^n \\
&\Longrightarrow a_0+\sum_{n\geq 1} \{a_nx^n-na_nx^n\}=0 \end{aligned}

Let $x=0$ to get $a_0=0$

Differentiate successively, and let $x=0$ to get $a_n-na_n=0 \quad \quad \forall n\geq 1$

Thus $a_n(n-1)=0 \quad \forall n\geq 1$

So either n=1, either $a_n=0$

So all the possible solutions are :
- $a_n=0 \quad \forall n\geq 0$ ---> $y=0$ is a solution

- $a_n=0 \quad \forall n\geq 2$ and $a_0=0$. And $a_1$ can be whatever you want. So $y=kx$ is a solution.

@ Tessy : you got the wrong equation

4. Thanks Moo I solve the wrong equation