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Math Help - Have I got this nonhomogeneous q right?

  1. #1
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    Have I got this nonhomogeneous q right?

    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...
    Last edited by 0renji; April 14th 2009 at 02:24 PM.
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  2. #2
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    Quote Originally Posted by 0renji View Post
    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.
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  3. #3
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    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.
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  4. #4
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    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}
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    Quote Originally Posted by danny arrigo View Post
    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.
    Last edited by 0renji; April 15th 2009 at 05:35 AM.
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