# Finding the series solution for a second order ode

Find the series solution for $y''-2y'+2y=0$

Assuming that $y=\Sigma^{\infty}_{n=0} c_nx^n$I got the recurrence relation: $c_nn(n-1)-2c_{n-1}(n-1)+2c_{n-2}=0$

Therefore: $c_3=\frac13c_1-\frac23c_0$
$c_4=-\frac16c_0$
$c_5=-\frac1{30}c_1$
$c_6=-\frac1{90}c_1+\frac1{90}c_0$
$c_7=-\frac1{630}c_1+\frac1{315}c_0$
$c_8=-\frac1{2520}c_0$,

Finding the underlying rule is just way beyond the limitation of my ability. What is the pattern of these expressions?

• First find the exact solution, which is of the form cexp[ax]. In fact there are two. Then expand the exponential. – Urgje Apr 12 '14 at 8:50
• No I just want to find out the pattern of these expressions – pxc3110 Apr 12 '14 at 8:51

The recurrence relation you are looking for is apparently $$\frac{c_0 \left((1-i)^{n+1}+(1+i)^{n+1}\right)+i c_1 \left((1-i)^n-(1+i)^n\right)}{2 n!}$$ which can be simplified to $$\frac{2^{n/2} \left(\sqrt{2} c_0 \cos \left(\frac{ \pi (n+1)}{4}\right)+c_1 \sin \left(\frac{\pi n}{4}\right)\right)}{n!}$$

• how did you get this please? – pxc3110 Apr 13 '14 at 5:05
• Almost magic ! I am joking, be sure. In fact,what gave me the idea is the explicit solution of the ODE. From there, everything becomes simple. – Claude Leibovici Apr 13 '14 at 5:07
• As I remarked in my earlier comment. This is a convenient way to obtain the solution. – Urgje Apr 13 '14 at 9:00
• @Urgje.You are perfectly right ! It helps a lot for getting some ideas. Cheers. – Claude Leibovici Apr 13 '14 at 9:02