# Y''' + 8y = 2x - 5 + 8 e^(-2x); y(0) = -5, y'(0) = 3, y''(0) = -4

1. Feb 20, 2014

### s3a

1. The problem statement, all variables and given/known data
The problem:
y''' + 8y = 2x - 5 + 8 e^(-2x); y(0) = -5, y'(0) = 3, y''(0) = -4

y = -23/12 e^(-2x) + e^x (-59/24 cos(sqrt(3) x) + 17 sqrt(3)/72 sin(sqrt(3) x)) + 1/4 x - 5/8 + 2/3 x e^(-2x)

2. Relevant equations
Method of undetermined coefficients.

3. The attempt at a solution
My work is attached as the TheProblemAndMyWork.pdf file. The work is typed (not handwritten), so it looks nice.

What I'm having trouble with is finding the values for c_1, c_2 and c_3. I entered the system of equations from my work into Wolfram Alpha, and the answer is wrong! I re-did this problem so many times, and I keep ending up in the same situation. I suppose, I keep making the exact same mistake.

Could someone please tell me what I am doing wrong?

Any help would be GREATLY appreciated!

#### Attached Files:

• ###### TheProblemAndMywork.pdf
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2. Feb 20, 2014

### vela

Staff Emeritus
The constant term you get when you calculated f''(0) should be -8/3, not 8.

3. Feb 20, 2014

### Ray Vickson

It is a quite a bit easier if you first get rid of the x and constant terms on the right. You can do this by looking at $z(x) = y(x) + ax + b$ for some constants $a, b$, then adjusting $a$ and $b$ until you get $z''' + 8z = 8 e^{ -2x}$. Then the "undetermined coefficient" equations simplify down quite a lot. Of course you need to determine the initial conditions on z(t) and use them instead.

4. Feb 22, 2014

### s3a

Thanks guys, I get it now!