Forcing series application

Question

$$\sum_{n=0}^∞ C_nx^n = {y(x)}{}$$

$ 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 :/

Answer 1

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 $y=A x^{i \sqrt{2}}+B x^{-i \sqrt{2}}$.

Answer 2

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