Have I got this nonhomogeneous q right?

**0renji** · Apr 14th 2009, 09:18 AM

Not sure about this one, a quick check would be appreciated and a bit of help as to where I've gone wrong if it's not right... a friend got a different answer so I'm questioning my own.

$y''(x) - 2y'(x) - 3y(x) = x^3$ subject to y(0) = 1 and y'(0) = -1

I got $y(x) = -e^{-x} + \frac{14}{27}e^{3x} - \frac{1}{3}x^3 + \frac{2}{3}x^2 - \frac{14}{9}x + \frac{40}{27}$

Testing y(0) there does get me 1 so it seems right to me...

**mr fantastic** · Apr 14th 2009, 02:10 PM

Originally Posted by 0renji

Not sure about this one, a quick check would be appreciated and a bit of help as to where I've gone wrong if it's not right... a friend got a different answer so I'm questioning my own.

$y''(x) - 2y'(x) - 3y(x) = x^3$ subject to y(0) = 1 and y'(0) = -1

I got $y(x) = -e^{-x} + \frac{14}{27}e^{3x} - \frac{2}{3}x^3 + \frac{2}{3}x^2 - \frac{14}{9}x + \frac{40}{27}$

Testing y(0) there does get me 1 so it seems right to me...

Your homogenous solution looks OK. Does the particular solution work when you substitute into the DE? If so, then you have the correct answer.

**0renji** · Apr 14th 2009, 02:37 PM

Just tried it out with x = 3 and indeed the result was 27 (I tried with 2 at first and the answer came out wrong but I think that was just a calculation mistake). I will take great delight in telling my friend he is wrong and I am right. Hahaha.

**Jester** · Apr 14th 2009, 03:38 PM

I think there's a problem here. If you want the solution subject to the following IC's, $y(0)=1,\;\;y'(0)=1$ , then the solution you give is right (with a slight change- should be $- \frac{1}{3}x^3$ ). However, if you want the solution with the IC's you give $y(0)=1,\;\;y'(0)=-1$ , I believe the correct solution is

$<br /> y(x) = -\frac{1}{2}e^{-x} + \frac{1}{54}e^{3x} - \frac{1}{3}x^3 + \frac{2}{3}x^2 - \frac{14}{9}x + \frac{40}{27}$

**0renji** · Apr 15th 2009, 05:21 AM

Originally Posted by danny arrigo

I think there's a problem here. If you want the solution subject to the following IC's, $y(0)=1,\;\;y'(0)=1$ , then the solution you give is right (with a slight change- should be $- \frac{1}{3}x^3$ ). However, if you want the solution with the IC's you give $y(0)=1,\;\;y'(0)=-1$ , I believe the correct solution is

$<br /> y(x) = -\frac{1}{2}e^{-x} + \frac{1}{54}e^{3x} - \frac{1}{3}x^3 + \frac{2}{3}x^2 - \frac{14}{9}x + \frac{40}{27}$

You're right, $-\frac{2}{3}x^3$ was a typo which I corrected in the first post eventually, and my y'(0) does come out to 1. Oops. No idea how I missed that. Thanks for the help, I'll go back and see where I went wrong when I worked it out...

edit: Ah, I missed a negative sign when working out y'(0)... stupid mistake heh. All fixed now.

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