I tried to solve $(x^2)y''+y=0$ using power series, but I cannot get the general solution or the relation at least
$$(x^2)y''+y=0 $$
$$ \sum_{n=2}^\infty c_n n(n-1) x^n + \sum_{n=0}^\infty c_n x^n$$
$$ \sum_{k=2}^\infty c_k k(k-1) x^k + \sum_{k=0}^\infty c_k x^k$$
$$ c_o+c_1+\sum_{k=2}^\infty c_k (k^2-k+1) x^k$$
From here $C_0=0$, $C_1=0$, $ C_k(k^2 -k+1)=0$.
Is this right? How can I get a recurrence relation for coefficients or the general solution of the series?
Best Answer
First rearrange $x^2y''+y=0$ to
$$y''+\frac{1}{x^2}y=0$$
Since the equation has a regular singular point a likely plan of attack assumes $y = \sum\limits_{n= 0}^\infty c_n x^{n+p}$ where $p$ is yet to be determined. Taking derivatives and substituting above gives
$$\sum\limits_{n= 0}^\infty (n+p)(n+p-1)c_n x^{n+p-2}+\frac{1}{x^2}\sum\limits_{n= 0}^\infty c_n x^{n+p}=0$$
$$\sum\limits_{n= 0}^\infty \bigg\{(n+p)(n+p-1) +1\bigg\} c_n x^{n+p-2}=0$$
For $n=0$, we have the lowest power of $x$. It's coefficient is $\big\{p(p-1)+1 \big\}c_0$. This must equal zero if the equation is to equal zero for any $x$. For a nontrivial solution, $c_0 \ne 0$. Then
$$\big\{p(p-1)+1 \big\}c_0=0$$ $$p^2-p+1=0$$
This has complex roots $p_{1,2} = \frac{1\pm i\sqrt{3}}{2}$, which implies that $y$ is a complex series where $c_n \in \mathbb{C}$. Substituting our roots back into the assumed form for $y$, this implies that the solutions have form
$$y = x^{p1}\sum\limits_{n= 0}^\infty c_n x^{n} \quad \mathtt{and} \quad y = x^{p2}\sum\limits_{n= 0}^\infty c_n x^{n}$$
There is a good deal of analysis that actually winds up with a generalized form for such a case! Solutions to the Frobenius equation with complex indicial roots are given by a general solution where the roots are $p_{1,2} = a +ib$:
$$y_1 = (x-x_0)^{a}\cos\left(b \ln|x-x_0| \right)$$ $$y_2 = (x-x_0)^{a}\sin\left(b \ln|x-x_0| \right)$$
where $x_0$ is the singular point. For us $x_0 = 0$, so for our roots we have
$$y_1 = A x^{1/2}\cos\left(\frac{\sqrt{3}}{2} \ln|x| \right)$$ $$y_2 = B x^{1/2}\sin\left(\frac{\sqrt{3}}{2} \ln|x| \right)$$
This matches WolframAlpha. I recommend researching the Frobenius Method where the indicial roots are complex for further information.