# Thread:

1. ## Find General Solution

Find the general solution of y'' - iy' + 6y = 0.

I know how to solve these, but the 'i' is throwing me off on this one...

Can someone show this one? Thanks a lot...

2. Have you found and solved the characteristic equation?

3. The first thing i did was write: r^2 - ir + 6 = 0
Then factored it to be (r+2i)(r-3i) = 0
So,
r = -2i and 3i

What from here? THanks.

4. Sounds like a good start. So you have complex solutions to your charateristic equation.

What general form should you use?

This example is really good to follow. Homogeneous linear equations with constant coefficients

5. The problem that I'm having is this:

a = 0, but b = -2 AND b = 3

So, would my general solution look something like this? What am I missing?
y(x) = e^0 (c1cos(-2) + c2sin(3)

6. Originally Posted by jzellt
The problem that I'm having is this:

a = 0, but b = -2 AND b = 3

So, would my general solution look something like this? What am I missing?
y(x) = e^0 (c1cos(-2) + c2sin(3)
you're close

as $e^0=1$ then $y(x) = c_1\cos(\beta x) + c_2\sin(\beta x)$

But now we have an additinal problem as the solutions aren't complex conjugates.

7. Exactly... Thats my problem. What do I do from here?

8. Originally Posted by jzellt
Exactly... Thats my problem. What do I do from here?
Since the DE is complex, why can't you just give $y = A e^{3ix} + B e^{-2ix}$ as the solution ....?

9. Yeah, not sure why I didn't realize that a first, but the important thing is that eventually I figured it out. Thanks

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