Forcing series application $$\sum_{n=0}^∞ C_nx^n = {y(x)}{}$$
$ x^2y''+xy'+2y=0 $
Can someone help me with this problem any soon please? I tried to solve it but I'm stucked.. I found $ x^n $ in all pieces. Thus, I can't go forward :/
 A: The general solution is $$y=a \cos \left(\sqrt{2} \log x\right)+b \sin \left(\sqrt{2} \log x\right)$$ which has no MacLaurin series and this is why you will not find a series expansion like $$y(x)=\sum _{k=0}^{\infty} c_k x^k$$
because first derivative is $$y'(x)=\sum _{k=1}^{\infty} kc_k x^{k-1}$$
second derivative is $$y''(x)=\sum _{k=2}^{\infty} k(k-1)c_k x^{k-2}$$
Plugging into the equation
$$x^2 y''(x)+x y'(x)+2 y(x)=0$$
we get
$$x^2\sum _{k=2}^{\infty} k(k-1)c_k x^{k-2}+x\sum _{k=1}^{\infty} kc_k x^{k-1}+2\sum _{k=0}^{\infty} c_k x^k=0$$
which is
$$\sum _{k=2}^{\infty} k(k-1)c_k x^{k}+\sum _{k=1}^{\infty} kc_k x^{k}+2\sum _{k=0}^{\infty} c_k x^k=0$$
which gives
$$\sum _{k=0}^{\infty} d_k x^k=0$$
where all $d_k=0$ for any $k$
So you need to expand in another point $x\ne 0$
hope this is useful
A: Assume there are solutions of the form $y=x^{ \lambda}$. So $y'=\lambda x^{ \lambda-1}$ and $y''=\lambda  (\lambda-1)x^{ \lambda-2}$ . Now substitute this into the differential equation
\begin{eqnarray*}
x^{ \lambda} (\lambda  (\lambda-1) +\lambda+2)=0.
\end{eqnarray*}
This gives $ \lambda = \pm  i \sqrt{2}$ . So the general solution is $y=A x^{i \sqrt{2}}+B x^{-i \sqrt{2}}$. 
