0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ First find the exact solution, which is of the form cexp[ax]. In fact there are two. Then expand the exponential. $\endgroup$ – Urgje Apr 12 '14 at 8:50
  • $\begingroup$ No I just want to find out the pattern of these expressions $\endgroup$ – pxc3110 Apr 12 '14 at 8:51
1
$\begingroup$

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 $$\frac{2^{n/2} \left(\sqrt{2} c_0 \cos \left(\frac{ \pi (n+1)}{4}\right)+c_1 \sin \left(\frac{\pi n}{4}\right)\right)}{n!}$$

$\endgroup$
  • $\begingroup$ how did you get this please? $\endgroup$ – pxc3110 Apr 13 '14 at 5:05
  • $\begingroup$ 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. $\endgroup$ – Claude Leibovici Apr 13 '14 at 5:07
  • $\begingroup$ As I remarked in my earlier comment. This is a convenient way to obtain the solution. $\endgroup$ – Urgje Apr 13 '14 at 9:00
  • $\begingroup$ @Urgje.You are perfectly right ! It helps a lot for getting some ideas. Cheers. $\endgroup$ – Claude Leibovici Apr 13 '14 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.