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

[s3a]

Homework Statement

The problem:
y''' + 8y = 2x - 5 + 8 e^(-2x); y(0) = -5, y'(0) = 3, y''(0) = -4

Correct answer:
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)

Homework Equations

Method of undetermined coefficients.

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!

 

[vela]

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

 

[Ray Vickson]

Homework Statement

The problem:
y''' + 8y = 2x - 5 + 8 e^(-2x); y(0) = -5, y'(0) = 3, y''(0) = -4

Correct answer:
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)

Homework Equations

Method of undetermined coefficients.

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!

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.

 

[s3a]

Thanks guys, I get it now!

 

[tatis ss]

Could someone explain to me how the exercise found in pdf is done, please thank you.

 

[vela]

That’s not how it works here. You need to show an attempt, and we can help you identify and correct your mistakes and misconceptions.

There should be plenty of examples you can see in your textbook.