# Non-conjugated roots in the eigenequation

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?

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

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.

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

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