# Homework Help: Differential Equations

1. Jan 7, 2006

### frozen7

Solve :

$$y'' + y' - 12y = 4x^2$$
The complementary equation I get is $$y1 = C1 e^3x + C2 e^-4x$$

But how to solve for the trial solution?
I do it in this way:

$$f(x) = 4x^2$$
$$y2 = D (Ax^2 + Bx + C )$$......
What I want to know is whether my y2 is correct.

Last edited: Jan 7, 2006
2. Jan 8, 2006

### Tide

Are you familiar with the method of variation of parameters?

3. Jan 8, 2006

### TD

You can certainly use undetermined coefficients since your RHS is just a polynomial. Since its degree is 2, your suggestion for a solution should be a general second degree polynomial as well.

Your y2 is fine, but it can be simplified a bit, the D isn't necessary. If you would work it out, you'd get ADx²+BDx+CD where 'AD', 'BD' and 'CD' are again just 3 constants so just using A, B and C is fine - then you have the most general second degree polynomial.

Find its first and second derivative, plug it into your DE and identify coefficients to solve for A, B and C

4. Jan 8, 2006

### mathmike

i think you should try variation of parameters as tide suggested

5. Jan 8, 2006

### frozen7

What is variation of parameters??

6. Jan 8, 2006

### TD

You can google it, the method is described well in http://www.math.utah.edu/~gustafso/2250variation-of-parameters.pdf" [Broken].
This is certainly a method worth learning since it applies more generally than the method of undetermined coefficients (which only works for a limited number of RHS functions)

Just as a note: your method (undetermined coefficients) will work here as well!

Last edited by a moderator: May 2, 2017
7. Jan 8, 2006

### frozen7

Thanks a lot.

8. Jan 8, 2006

### TD

You're welcome, don't hesitate to ask for help if you're stuck