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Homework Help: Diff Eq: Is it me? Or Wolfram? I know, it's me.

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data

    y(5) - 2y(4) + y(3) = 0

    This is 5th-order, linear, HG, with constant coefficients, and hence I used the characteristic equation:

    m5 - 2m4 + m3 =

    m3(m - 1)(m - 1) = 0

    Hence, m = {1,1,0,0,0} and so the solution is given by:

    y(x) = C1 + C2*x +C3*x2 + C4*exp(x) + c5*x*exp(x).

    This is the Wolfram solution.

    They seem to have an extra exponential term inside of the parentheses. Am I missing something?

    Thanks!
     
  2. jcsd
  3. Sep 27, 2010 #2
    Hmmm.... looking at the characteristic EQ, it would seem that:

    m3(m2 - 2m + 1) = 0 and hence:

    (m2 - 2m + 1) = 0 and so

    y'' - 2y' + y = 0

    But when you look at my solution, differentiation will cause my C1 term to drop out of the derivatives altogether, and hence it will not be able to cancel.

    I have messed up somehow, but cannot seem to rectify it. I have tunnel vision right now and cannot seem to convince myself that my method is not correct.
     
  4. Sep 27, 2010 #3

    cristo

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    Doesn't the wolfram solution reduce to yours on setting their c_1-3c_2 equal to your C_4?
     
  5. Sep 27, 2010 #4

    rock.freak667

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    In Wolfram's answer, you can just write c1-3c2 as some other symbol and get your same answer. But I am not sure exactly how Wolfram solved it.
     
  6. Sep 27, 2010 #5
    I am still not seeing how that rectifies this. Maybe I need a coffee ...
     
  7. Sep 27, 2010 #6

    rock.freak667

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    Your solution would not solve y'' - 2y' + y = 0

    it will solve d3/dx3 (y'' - 2y' + y) to give 0.
     
  8. Sep 27, 2010 #7

    D H

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    Your solution (I changed c5 to C5):

    [tex]y(x) = C_1 + C_2x +C_3x^2 + C_4\exp(x) + C_5x\exp(x)[/tex]

    Wolfram's solution:

    [tex]y(x)=c_5x^2 + c_4x+e^x(c_2x+c_1-3c2)+c3[/tex]

    Those solutions are the same. To see this, set

    [tex]\aligned
    C_1 &= c_3 \\
    C_2 &= c_4 \\
    C_3 &= c_5 \\
    C_4 &= c_1 - 3c_2 \\
    C_5 &= c_2
    \endaligned[/tex]
     
  9. Sep 27, 2010 #8
    Ok! Thanks guys :smile:

    Hmmm.

    [tex]\frac{d^5y}{dx^5} - 2\frac{d^4y}{dx^4} + \frac{d^3y}{dx^3} = 0[/tex]

    [tex]\Rightarrow \frac{d^3y}{dx^3}\left(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} +1\right) = 0[/tex]

    Ok. I see that it wil not solve y'' - 2y' + y = 0. But from the above, should it not solve

    [tex]\left(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} +1\right) = 0[/tex]

    which does not look promising. I know that I am seriously abusing the differential operator, but it should work eh?
     
  10. Sep 27, 2010 #9

    rock.freak667

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    y'' - 2y' + y = 0

    has a solution of y=(Ax+B)e-x, which will solve y'' - 2y' + y = 0

    Your general solution which contains more terms, I do not think will solve y'' - 2y' + y = 0.
     
  11. Sep 27, 2010 #10
    Ah ha. I see. I was trying to plug in my general solution to

    [tex]
    \frac{d^5y}{dx^5} - 2\frac{d^4y}{dx^4} + \frac{d^3y}{dx^3} = 0
    [/tex]

    which will of course have more terms than the general solution to just

    [tex]
    \left(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} +1\right) = 0
    [/tex]

    Thanks again! :smile:
     
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