Solving $x^2y''+2xy'+2y=0$ using power series

Question

One way to solve $x^2y''+2xy'+2y=0$ is using the substitution $x=e^t$. This time I'm asked to use power series (Fröbenius method). Assuming $y=\sum_{n\geq 0}a_nx^{n+r}$, the diff eq reduces to $$\sum_{n\geq 0}(n+r)(n+r-1)a_nx^{n+r}+\sum_{n\geq 0}2(n+r)a_nx^{n+r}+\sum_{n\geq 0}2a_nx^{n+r}=0$$ $$\sum_{n\geq 0}\{(n+r)(n+r-1)+2(n+r)+2\}a_nx^{n+r}=0$$

This is the part where I'm lost at:

We need to impose $(n+r)(n+r-1)+2(n+r)+2=0$ as $a_n\neq 0$, right? Defining $u\equiv n+r$ we get $$u(u-1)+2r+2=u^2+u+2=0\implies u=n+r=-\frac{1}{2}\pm i\frac{\sqrt 7}{2}$$ So I'm guessing the solution should look like $$y=\sum_{n\geq 0}a_nx^{-\frac{1}{2}+ i\frac{\sqrt 7}{2}}+\sum_{n\geq 0}a_nx^{-\frac{1}{2}- i\frac{\sqrt 7}{2}}$$ But what about the $a_n$ s? What can be said about them? In view that the differential equation is a $2$nd order one, shouldn't there be only two coefficients $a_0$ and $a_1$ determined by boundary conditions? Therefore, somehow, $$y=a_0x^{-\frac{1}{2}+ i\frac{\sqrt 7}{2}}+a_1x^{-\frac{1}{2}- i\frac{\sqrt 7}{2}}=\frac{1}{\sqrt x}\left(a_0e^{i\frac{\sqrt 7}{2}\ln x}+a_1e^{-i\frac{\sqrt 7}{2}\ln x}\right)=\frac{A}{\sqrt x}\sin\left(\frac{\sqrt 7}{2}\ln x+B\right)$$

Answer 1

Let's call the indicial roots $r_+ = -1/2 + i \sqrt{7}/2$ and $r_- = -1/2 + i \sqrt{7}/2$. Your solutions (for $x > 0$) are $$y = a x^{r_{+}} + b x^{r_{-}} = \frac{a}{\sqrt{x}} \exp(i \sqrt{7} \ln(x)/2) + \frac{b}{\sqrt{x}} \exp(-i \sqrt{7} \ln(x)/2)$$ Now use the formulas $\exp(it) = \cos(t) + i \sin(t)$ and $\sin(t + B) = \sin(t) \cos(B) + \cos(t) \sin(B)$ to express $\sin(\sqrt{7} \ln(x)/2+B)/\sqrt{x}$ in this form.

Answer 2

When you see groups $x^ky^{(k)}$ where the power for $x$ is the same as the order of derivation for $y$ then you are dealing with Cauchy-Euler ODE.

The recommended substitution in this case is $y(x)=u(\ln(x))$

You will get a linear one with constant coefficients $$u''+u'+2u=0$$

Characteristic equation $r^2+r+2=0\iff r=-\frac 12\pm i\frac{\sqrt{7}}{2}$

$\begin{align}y(x)&=e^{-\frac 12\ln(x)}\Big(A\cos(\frac{\sqrt{7}}2\ln(x))+B\sin(\frac{\sqrt{7}}2\ln(x)\Big)\\\\&=\frac{A}{\sqrt{x}}\cos\left(\frac{\sqrt{7}}2\ln(x)\right)+\frac{B}{\sqrt{x}}\sin\left(\frac{\sqrt{7}}2\ln(x)\right)\end{align}$

Which you can transform to amplitude and phase instead if you wish.

Using Frobenius method here will mostly be difficult because the resulting series will be a composition of series for $\sin,\cos$ and $\ln$ so it won't be trivially recognizable.