0
$\begingroup$

I have an ODE:

$$y'' - iky = 0, k > 0$$

I tried to solve it via the eigenequation, and the result is $r^2 = ik$, $r = \pm k(\sqrt{2}/2 + i\sqrt{2}/2)$

And the roots are non-conjugated...

What should I do?

$\endgroup$
  • $\begingroup$ Conjugate roots appear when the equation has real coefficients. This does not prevent you to solve the ODE. $\endgroup$ – Siméon Oct 2 '13 at 6:20
  • $\begingroup$ @ShawnWang: Did the answer resolve your issue? $\endgroup$ – Amzoti Oct 3 '13 at 18:31
  • $\begingroup$ @Ju'x Thanks! For both editing and commenting. $\endgroup$ – Shawn Wang Oct 4 '13 at 2:42
1
$\begingroup$

Hint:

Assume the solution is:

$$y(t) = e^{m t}$$

Find the second derivative, substitute into $(1)$ and solve.

Spoiler Hover over the following area.

$ \displaystyle y(t) = y_1(t) + y_2(t) = c_1 e^{-(1+i) \sqrt{k} t/\sqrt{2}} + c_2 e^{(1+i) \sqrt{k} t/\sqrt{2}} $

$\endgroup$
  • $\begingroup$ The spoiler area is grey for me... $\endgroup$ – lhf Oct 3 '13 at 4:42
  • $\begingroup$ @Amzoti Thank you! I used a same equation that y = c1 * exp (r1*x) + c2 * exp (r2*x) :) $\endgroup$ – Shawn Wang Oct 4 '13 at 2:43
  • $\begingroup$ @ShawnWang: You are very welcome! Regards $\endgroup$ – Amzoti Oct 4 '13 at 2:45
  • $\begingroup$ How has this helpful post gone without a TU! Need to fix that! +1 $\endgroup$ – Namaste Oct 4 '13 at 12:57
  • $\begingroup$ Yes it is POETS Day! One week down for me, one to go at L'Hospital! $\endgroup$ – Namaste Oct 4 '13 at 12:58

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.