Question

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

Answer 1

Problem:

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

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

Or:

$(a_{0}x^{\sigma}\times\sigma\times(\sigma-1) +a_{1}x^{\sigma+1}\times\sigma\times(\sigma+1)+a_2x^{\sigma+2}\times(\sigma+1)\times(\sigma+2)+...) + (a_0x^{\sigma+1}\times\sigma+a_1x^{\sigma+2}\times(\sigma+1)+a_2x^{\sigma+3}\times(\sigma+2)+...) - (a_0x^\sigma\times\sigma+a_1x^{\sigma+1}\times{\sigma+1}+a_2x^{\sigma+2}\times(\sigma+2)+...) + (a_0x^{\sigma}+a_1x^{\sigma+1}+a_2x^{\sigma+2}+...)-(a_0x^{\sigma+1}+a_1x^{\sigma+2}+a_2x^{\sigma+3}+...)=0$

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

$\begin{cases} a_0\times\sigma\times(\sigma-1)-a_0\times\sigma+a_0=0\\ a_1\times\sigma\times(\sigma+1)+a_0\times\sigma-a_1\times(\sigma+1)+a_1-a_0=0\\ a_2\times(\sigma+1)\times(\sigma+2)+a_1\times(\sigma+1)-a_2\times(\sigma+2)+a_2-a_1=0\\ ... \end{cases}$

From the first equation (we take that $a_0\not=0$ )we get: $(\sigma-1)^2=0=>\sigma=1$

Let's substitute this value to our system (3). Now we have:

$\begin{cases} a_0\not=0 \\ a_1=0 \\ a_2=0\\ ... \end{cases}$

So we have that all coefficient $a_k$ (except $a_0$ ) are equal zero. That means that:

$y=a_0x^\sigma$

Also we know that $\sigma=1$. So solution of this problem is:

$y=a_0x$,

where $a_0$ is a different constant.