Question

2x(1-x)y"+(1-x)y'+3y=0 solve by forbeinus series.

Answer 1

Here $P=\frac{1}{2x}; Q=\frac{1}{2x(1-x)}$

THus x=0 and x=1 are the regular points of the equation

Let's find the solution in form

$y=\sum\limits_{n=0}^{\infty}a_nx^{n+r}$

Hence

$y'=\sum\limits_{n=1}^{\infty}(n+r)a_nx^{n+r-1}=\sum\limits_{n=0}^{\infty}(n+r+1)a_{n+1}x^{n+r}$

$y''=\sum\limits_{n=1}^{\infty}(n+r)(n+r+1)a_{n+1}x^{n+r-1}=\sum\limits_{n=0}^{\infty}(n+r+1)(n+r+2)a_{n+2}x^{n+r}$

Plugging it into the equation, we get

$2x(1-x)\sum\limits_{n=0}^{\infty}(n+r+1)(n+r+2)a_{n+2}x^{n+r}+(1-x)\sum\limits_{n=0}^{\infty}(n+r+1)a_{n+1}x^{n+r}+3\sum\limits_{n=0}^{\infty}a_nx^{n+r}=0$

$\sum\limits_{n=0}^{\infty}2(n+r+1)(n+r+2)a_{n+2}x^{n+r+1}-\sum\limits_{n=0}^{\infty}2(n+r+1)(n+r+2)a_{n+2}x^{n+r+2}+$

$+\sum\limits_{n=0}^{\infty}(n+r+1)a_{n+1}x^{n+r}-\sum\limits_{n=0}^{\infty}(n+r+1)a_{n+1}x^{n+r+1}+3\sum\limits_{n=0}^{\infty}a_nx^{n+r}=0$

$\sum\limits_{n=1}^{\infty}2(n+r+1)(n+r)a_{n+1}x^{n+r}-\sum\limits_{n=2}^{\infty}2(n+r-1)(n+r)a_{n}x^{n+r}+$

$+\sum\limits_{n=0}^{\infty}(n+r+1)a_{n+1}x^{n+r}-\sum\limits_{n=1}^{\infty}(n+r)a_{n}x^{n+r}+3\sum\limits_{n=0}^{\infty}a_nx^{n+r}=0$

For $x^r$ we get

$(r+1)a_{1}+3a _0=0$

For $x^{r+1}$

$2(r+2)(r+1)a_2+(r+2)a_1-(r+1)a_1+3a_1=0$