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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?

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  • $\begingroup$ Put your work in the question. The comments are not part of the question. (For one thing, people who know how to format math for this site can't format your comment so others can read it...) $\endgroup$ – Thomas Andrews Nov 21 '15 at 3:18
  • $\begingroup$ What is the question? $\endgroup$ – copper.hat Nov 21 '15 at 3:41
  • $\begingroup$ First, are my solution correct???, if right? I cannot formulate recurrence relation, then I cannot find C3,C4...... to formulate close expression for the general solution of power series. $\endgroup$ – AHT Nov 21 '15 at 3:49
  • $\begingroup$ Whoa... You've got $x,X,k,K,$and $N$ to boot. Be consistent. You can write subscripts with the underscore like c_0 to $c_0$. Also your fourth line actually implies that $c_0 + c_1 = 0 \implies c_0 = -c_1$. And since the equation is singular at $x=0$ when written in general form, perhaps a Frobenius series may give you the actual solution, not a standard pwoer series. $\endgroup$ – zahbaz Nov 21 '15 at 4:11
  • $\begingroup$ Just a minor detail : the expansion would be $c_0+c_1x+\sum_{k=2}^\infty c_k (k^2-k+1) x^k=0$ which leads to all $c_i=0$. This means that you cannot expand as series. $\endgroup$ – Claude Leibovici Nov 21 '15 at 4:31
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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.

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  • $\begingroup$ I was just writing in the same direction ! Good shot. Cheers. $\endgroup$ – Claude Leibovici Nov 21 '15 at 4:57
  • $\begingroup$ @ClaudeLeibovici. Hah thanks. I gave in to the general solution when I realized the roots were complex, though! I figure a textbook explanation is the obvious next step. $\endgroup$ – zahbaz Nov 21 '15 at 5:05
  • $\begingroup$ thanks guys for your valuable help. $\endgroup$ – AHT Nov 21 '15 at 14:25

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