Question

(x^2+1) y"+xy'-y=0 solve by power series

Answer 1

Problem:

$(x^2+1)\times \dfrac{\partial^2 y}{\partial x^2}+x\times \dfrac{\partial y}{\partial x}-y=0$ (1)

Let's image solution of the task in the next view:

$y=x^\sigma(a_{0}+a_{1}x+a_{2}x^2+...a_{k}x^k+...)$ (2)

Now subsitute equation (2) into equation (1):

$(x^2+1)\times(\sigma\times(\sigma-1)\times a_{0}x^{\sigma-2}+(\sigma+1)\times\sigma\times a_{1}x^{\sigma-1}+(\sigma+2)\times(\sigma+1)\times a_{2}x^{\sigma}+...)+x\times(\sigma\times a_{0}x^{\sigma-1}+(\sigma+1)\times a_{1}x^{\sigma}+(\sigma+2)\times a_{2}x^{\sigma+1}+...)-(a_{0}x^{\sigma}+a_{1}x^{\sigma+1}+a_{2}x^{\sigma+2}+...)=0$

Or:

$\sigma\times(\sigma-1)\times a_{0}x^{\sigma-2} + (\sigma+1)\times\sigma\times a_{1}x^{\sigma-1} + (\sigma\times(\sigma-1)\times a_{0}+(\sigma+2)\times(\sigma+1)\times a_{2}+\sigma\times a_{0}-a_{0})x^{\sigma}+...=0$

Now equate all coefficient near $x^{\sigma-2}$ , $x^{\sigma-1}$, $x^{\sigma}$,... to zero:

$\begin{cases} a_{0}\times(\sigma-1)\times\sigma=0 \\ a_{1}\times(\sigma+1)\times\sigma=0 \\ a_{0}\times(\sigma+1)\times(\sigma-1)-a_{2}\times(\sigma+2)\times(\sigma+1)=0 \\ a_{1}\times\sigma\times(\sigma+2)-a_{3}\times(\sigma+3)\times(\sigma+2)=0\\ .... \\ a_{k}\times(\sigma-k+1)\times(\sigma+k+1)-a_{k+2}\times(\sigma+k+2)\times(\sigma+k+1)=0\\ \end{cases}$ (3)

From the first equation (using that $a_{0}\not=0$) we get:

$\sigma=0$ or $\sigma=1$

Case: $\sigma=1$

Our system (3) will have the next view:

$\begin{cases} 0=0\\ a_{1}\times 2=0\\ a_{0}\times 0-a_{2}\times 3=0\\ a_{1}\times 3-a_{3}\times12=0\\ ... \end{cases}$

Or:

$\begin{cases} 0=0\\ a_{1}=0\\ a_{2}=0\\ a_{3}=0\\ ...\\ \end{cases}$

So we have that all coefficient $a_{k}$ (except $a_{0}$) are equal zero. So solution of this problem in this case is:

$y=a_{0}\times x^{\sigma}$,

But for this case $\sigma=1$, so

$y=a_{0}\times x$, where $a_{0}$ is a different constant.

Case: $\sigma=0$

Our system (3) will have the next view:

$\begin{cases} 0=0\\ 0=0\\ -a_{0}-2\times a_{2}=0\\ a_{1}\times 0 - a_{3}\times 6=0\\ ... \end{cases}$

Or:

$\begin{cases} a_{2}=-\dfrac{a_{0}}{2}\\ a_{3}=0\\ ... \end{cases}$

So we have that coefficient $a_{3}$, $a_{5}$, ... are equals to zero, and the pairs coefficient(and $a_{1}$) aren't equal to zero. For convenience let's get: $a_{1}=0$.

For this case, the genral solution will have the next view:

$y=a_{0}+a_{2} x^2+ a_{4}x^4+...= \begin{matrix} \\ \sum \\ _{k=0} \end{matrix} a_{2k}x^{2k}$