# Using Undetermined Coefficients to solve an equation for a particular solution?

1. Oct 18, 2014

### KevinD6

1. The problem statement, all variables and given/known data
$y'' + 9y = 3sin(3x) + 3 + e^{3x}$

2. Relevant equations

3. The attempt at a solution
This is my first post here so let me know if I've done anything wrong, I've been looking at questions here for a long time though ^^.

So the problem asks me to solve for one particular solution by using undetermined coefficients, I begin by solving for the homogeneous solution:
y'' + 9y = 0
The characteristic polynomial becomes:
$r^2 + 9 = 0$
Therefore, I get the two roots ${0 \pm 3i}$

Solving for the homogeneous solution, which I'll call $y_h$ I get:
$y_h = cos(3x) + sin(3x)$

Now, my professor hasn't gone through undetermined coefficients in his class yet but put it on the assignment (weird right?), so this is the part I might have screwed up on due to lack of knowledge.

Based on what I've found on the internet, first I predict a form of the particular solution, which I will denote $y_p$
So:
$y_p = A(cos(3x)) + B(sin(3x)) + C + D(e^{3x})$
Afterwards, I plug this particular solution back into my original equation in order to solve for the coefficients A, B, C, and D.

However, taking the second derivative of my particular equation, I get:
$y'' = -9A cos(3x) - 9B sin(3x) + 9D e^{3x}$
That equation is added onto 9y, which is:
$9y = 9Acos(3x) + 9B sin(3x) + 9C + 9D e^{3x}$

I'm left with the final equation of:
$9C + 18D = 3sin(3x) + 3 + e^{3x}$

Now what I'm confused about is the fact that A and B cancel out in this case, and I've been looking at internet explanations of undetermined coefficients but I can't find an answer to this question. Can anyone help me?

2. Oct 18, 2014

### LCKurtz

$\cos(3x)$ and $\sin(3x)$ satisfy the homogeneous equation so they can't contribute to the non-homogeneous. That's why they are dropping out.

Use $Ax\cos(3x) + Bx\sin(3x)$ in your trial solution for $y_p$.

3. Oct 18, 2014

### HaLAA

4. Oct 18, 2014

### HaLAA

x(Acos3x+Bsin3x)+C+De^3x

5. Oct 19, 2014

### KevinD6

Ahh, I see, it was just my ignorance about this method. Thanks guys! I hope to be posting here more in the future.

6. Oct 19, 2014

### GFauxPas

Paul's math notes has a good explanation of this method of multiplying throughout by a variable fixes problems sometimes.