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Solve 2nd-order differential eqn

  1. Jan 31, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    [tex]2y''-3y'+2y=x^3-5x+2[/tex]
    The attempt at a solution
    I recognized the R.H.S. as a polynomial of degree 3, so i have to expand up to [itex]D^3[/itex]

    The answer i got is quite close to that in my notes, except for one coefficient, so i was hoping someone could check it. It's a very long problem so i don't know if i should type all the steps here from my copybook. Please let me know.

    The answer in my notes:
    http://s1.ipicture.ru/uploads/20120201/pdG4AE7V.png
    The answer that i got is a different coefficient for x. So, i got 4x instead of 10x, but everything else is exactly as the answer that i'm supposed to get. I checked my long calculations and it appears to be correct, although it's a nagging feeling to be unsure about whether my entire method is flawed or not.
     
  2. jcsd
  3. Jan 31, 2012 #2

    Dick

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    If the only difference is 4x versus 10x your method is probably correct. It's probably just a numerical error. I did check that 10x works. 4x doesn't.
     
  4. Jan 31, 2012 #3

    sharks

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    OK, thanks for checking. I will go over my calculations again...
     
    Last edited: Jan 31, 2012
  5. Jan 31, 2012 #4

    sharks

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    I had made a small mistake. I got the answer 10x. However, i'm curious... How did you check that 10x works? Did you just differentiate the entire answer twice and then plug y, y' and y'' into the problem in post #1? Or is there an easier way of checking if the answer is correct?
     
  6. Feb 1, 2012 #5
    Yes. That is how you check.
     
  7. Feb 1, 2012 #6

    Dick

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    Well, I put A=0 and B=0 before I checked. I was just checking the inhomogeneous part.
     
  8. Feb 2, 2012 #7

    sharks

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    That's a nice trick, Dick. I understand the principle behind putting A=0 and B=0.

    But how can [itex]x^3-5x+2[/itex] be equivalent to [itex]\frac{1}{8}(4x^3+18x^2+10x-13)[/itex]?
     
  9. Feb 2, 2012 #8
    Why do you think it should be? You have [itex]y_p=\frac{1}{8}(4x^3+18x^2+10x-13)[/itex], so what does [itex]2y_p''-3y_p'+y_p = ?[/itex]
     
  10. Feb 2, 2012 #9

    sharks

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    OK, [itex]y_p=\frac{x^3-5x+2}{2D^2-3D+2}[/itex] which is equivalent to [itex]\frac{1}{8}(4x^3+18x^2+10x-13)[/itex]?

    But it's still not so easy to check the inhomogeneous part with actually computing [itex]y_p=\frac{x^3-5x+2}{2D^2-3D+2}[/itex] or maybe i'm wrong?
     
  11. Feb 2, 2012 #10

    Dick

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    To CHECK the inhomogeneous part all you have to do is operate with [itex]2D^2-3D+2[/itex] on [itex]\frac{1}{8}(4x^3+18x^2+10x-13)[/itex] to make sure you get [itex]x^3-5x+2[/itex]. That's easier than doing the reverse to find [itex]y_p[/itex] to begin with.
     
  12. Feb 2, 2012 #11

    sharks

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    Thanks for the hint, Dick. I verified your method and it's indeed easier to check that way.
     
  13. Feb 2, 2012 #12
    You still have not shown any work. I'm not sure what the problem is:

    [itex] y_p=\frac{4}{8}x^3+\frac{18}{8}x^2+\frac{10}{8}x-\frac{13}{8} [/itex]

    [itex] y_p'=\frac{12}{8}x^2+\frac{36}{8}x+\frac{10}{8} [/itex]

    [itex] y_p''=\frac{24}{8}x+\frac{36}{8} [/itex]

    and so

    [itex]2y''_p-3y_p'+2y = \frac{1}{8}(x^3-5x+2)[/itex].
     
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