Find the power series solution for the equation $\frac{dy}{dx} = 2xy$ The given is 
$$\frac{dy}{dx} = 2xy$$
Then moving everything to one side and changing the terms
$$y^{'}-2xy =0 $$
We assume that $y= \sum^\infty_{n=0}C_nx^n$ and taking the derivative is:
$$y^{'}=\sum^\infty_{n=1}nC_nx^{n-1}$$
Then plugging back into the given yields:
$$\sum^\infty_{n=1}nC_nx^{n-1}-\sum^\infty_{n=0}2nC_nx^{n+1}=0$$
Now we need to reindex:
$$C_1+\sum^\infty_{k=0}\left[(k+1)C_{k-1}-2C_{k+1}\right]x^{k}=0$$
Then we can see that $C_1=0$ and our recurrence formula is:
$$C_{k+1}=\frac{2C_{k-1}}{k+1}$$
Then after finding the different values of $k$ in the recurrence gives us:
$$k=1:C_2=\frac{2C_0}{2}=C_0$$
$$k=2:C_3=\frac{2C_1}{3}$$
$$k=3: C_4=\frac{2C_2}{4}$$
So our final answers are:
$$y_1=C_0$$
$$y_2=\sum^\infty_{n=1} \frac{x^{2n}}{n!}$$
Was I correct in my evaluation of the series?
 A: Clearly $y_1=C_0$ is not a solution and you forgot to multiply the terms in $y_2$ by $C_0$ that is not a solution also.
Following your own computations, the solutions are $$y(x)=y_1+C_0y_2=C_0\sum^\infty_{n=0} \frac{x^{2n}}{n!}.$$
A: $$\sum^\infty_{n=1}nC_nx^{n-1}-\sum^\infty_{n=0}2C_nx^{n+1}=0$$
the next step is not correct. It should be:
$$\sum^\infty_{n=0}(n+1)C_{n+1}x^{n}-\sum^\infty_{n=1}2C_{n-1}x^{n}=0$$
$$C_1+\sum^\infty_{n=1}(n+1)C_{n+1}x^{n}-\sum^\infty_{n=1}2C_{n-1}x^{n}=0$$
$$C_1+\sum^\infty_{n=1}((n+1)C_{n+1}-2C_{n-1})x^{n}=0$$
Hence  for $n \ge 1$:
$$C_1=0  \\
C_{n+1}=\dfrac {2C_{n-1}}{n+1}
$$
The recurrence formula is:
$$C_{2n}=\dfrac {C_0}{n!}$$
Therefore the solution is:
$$y(x)=C_0\sum_{n=0}^\infty \dfrac {x^{2n}}{n!}=C_0e^{x^2}$$
A: If I may suggest, never change the index
$$y= \sum^\infty_{n=0}C_n\,x^n \implies y'=\sum^\infty_{n=\color{red}{0}}n\,C_n\,x^{n -1}$$
$$y'-2xy =0\implies \sum^\infty_{n=0}n\,C_n\,x^{n -1}-2\sum^\infty_{n=0}C_n\,x^{n+1}=0$$ 
So, to have degree $m$ in the first sum, you must make $n-1=m$ that is to say $n=m+1$ and to have degree $m$ in the second sum, you must make $n+1=m$ that is to say $n=m-1$. Then
$$(m+1)\, C_{m+1}-2\, C_{m-1}=0\implies C_{m+1}=\frac{2\, C_{m-1}}{m+1}$$
