Hi, I am having trouble solving this problem, please help me figure this one out. Thanks in advance.
xy'' + y' + xy=0 at ordinary point x=1.
With the substitution $\displaystyle x-1=\xi$ the equation becomes...
$\displaystyle (1+\xi)\cdot y^{''} + y^{'} + (1+\xi)\cdot y=0$ (1)
That is an 'incomplete' linear ODE and we will search an analytic solution written as...
$\displaystyle y(\xi)= \sum_{n=0}^{\infty} a_{n}\cdot \xi^{n}$ (2)
If we derive from (2) le derivatives of $\displaystyle y(\xi)$ and subsitute them in (1) we arrive to write the following 'infinite sysytem' of algebric equations...
$\displaystyle a_{n-3} + a_{n-2} + n\cdot (n-1)\cdot (a_{n-1} + a_{n})=0$ (3)
... whose solution is...
$\displaystyle a_{n}= -a_{n-1} - \frac{ a_{n-2} + a_{n-3}}{n\cdot (n-1)}$ (4)
Since a solution of (1) multiplied by a constant is also a solution of (1), we can set without limitations $\displaystyle a_{0}=1$ and so with (4) we derive...
$\displaystyle a_{1}= -1, a_{2}= \frac{1}{2}, a_{3}= -\frac{1}{2}, a_{4}= \frac{13}{24}, a_{5}= - \frac{13}{24}, ... $
The searched solution od (1) is then...
$\displaystyle y(x)= 1 - (x-1) + \frac{1}{2}\cdot (x-1)^{2} - \frac{1}{2}\cdot (x-1)^{3} + \frac{13}{24}\cdot (x-1)^{4} - \frac{13}{24}\cdot (x-1)^{5} + ... $ (5)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$