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Math Help - Higher Order ODE IVP

  1. #1
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    Higher Order ODE IVP

    Hey MHF,

    I got stuck on finding the gen sol. to the following ODE

    y^{(4)}+4y = 0

    I solved the auxiliary eqn. m^4+4= 0 for m= ^4\sqrt{-4}

    So the roots must be complex? This is where I got stuck.
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  2. #2
    Super Member Random Variable's Avatar
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    The four roots are 1 \pm i and  -1 \pm i.

    So the general solution is  y(x) =  C_{1} e^x \cos x + C_{2} e^{x} \sin x + C_{3} e^{-x} \cos x + C_{4}e^{-x} \sin x
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  3. #3
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    Ok so I can use that to solve the IVP.

    y(0)=0 so 0 = C1 + C3
    and
    y'(x)= C1e^x(cosx-sinx)+C2e^x(sinx+cosx)-C3e^{-x} (cosx+sinx)+C4e^{-x}(cosx-sinx)
    y'(0) = 1 so
    1=C1+C2-C3+C4

    Is this correct so far?
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  4. #4
    Super Member Random Variable's Avatar
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    Are there any other conditions?
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  5. #5
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    Yes but I figured since it only gets messier i'd make sure I got this much correct. The other conditions are:

    y''(0)=0 and
    y'''(0)=2

    I just do what I did above to find the constants?
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  6. #6
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    If what I did above is correct then
    y''(x)= 2C2e^xcosx-2C1e^xsinx+2C3e^{-x}sinx-2C4e^{-x}cosx then
    0=2C2-2C4

    y'''(x)= 2C2e^x(cosx-sinx)-2C1e^x(sinx+cosx)+2C3e^{-x} (cosx-sinx)-2C4e^{-x}(cosx+sinx) and
    2= 2C2-2C1+2C3-2C4

    Idk if i'm on the right track here...
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  7. #7
    Super Member Random Variable's Avatar
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    There is an error somewhere because the coefficients should be  C_{1}=0, C_{2}= \frac{1}{2}, C_{3} = 0, C_{4} = \frac{1}{2} , and I'm not getting that by solving the four equations simultaneously.
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  8. #8
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    I know y' is correct, I will double check y'' and y'''. Those roots make all four equations true though, how did you find them without solving the four equations simultaneously? Thanks.
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  9. #9
    Super Member Random Variable's Avatar
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    Quote Originally Posted by mmattson07 View Post
    I know y' is correct, I will double check y'' and y'''. Those roots make all four equations true though, how did you find them without solving the four equations simultaneously? Thanks.
    I solved the problem using Maple.

    Give me a second to solve the problem myself by hand and see what I get.
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  10. #10
    Super Member Random Variable's Avatar
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     y(x) =  C_{1} e^x \cos x + C_{2} e^{x} \sin x + C_{3} e^{-x} \cos x +  C_{4}e^{-x} \sin x

     y'(x) = C_{1} (e^{x} \cos x - e^{x} \sin x)  + C_{2} (e^{x} \sin x + e^{x}  \cos x) + C_{3} (-e^{-x} \cos x - e^{-x} \sin x) + C_{4}  (-e^{-x} \sin x+ e^{-x} \cos x)

     y''(x) = C_{1}(e^{x} \cos x -e^{x} \sin x -e^{x} \sin x - e^{x} \ \cos x) + C_{2}( e^{x}\sin x + e^{x}\cos x + e^{x} \cos x - e^{x} \sin x) + C_{3} (e^{-x}\cos x + e^{-x} \sin x+ e^{-x} \sin x - e^{-x} \cos x) + C_{4}  ( e^{-x} \sin x - e^{-x} \cos x -e^{-x} \cos x - e^{-x} \sin x)

     = -2C_{1}(e^{x}\sin x) + 2 C_{2}e^{x}\cos x + 2C_{3}e^{-x}\sin x -2C_{4} e^{-x} \cos x

     y'''(x) = -2C_{1}(e^{x} \sin x + e^{x} \cos x) + 2C_{2}(e^{x} \cos x - e^{x} \sin x) + 2C_{3} (-e^{-x} \sin x +e^{-x} \cos x) - 2C_{4} (-e^{-x}\cos x - e^{-x} \sin x)

    y(0) = 0 = C_{1}+ C_{3}

     y'(0) = 1= C_{1} + C_{2} - C_{3} + C_{4}

    EDIT:  y''(0) = 0 = 2C_{2} - 2C_{4}

     y'''(0) = 2 = -2 C_{1} + 2C_{2} +2C_{3} + 2C_{4}

    which when solved, gives  C_{1}=0, C_{2} = \frac{1}{2}, C_{3} =0, C_{4} = \frac{1}{2}
    Last edited by Random Variable; March 20th 2010 at 12:12 PM.
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  11. #11
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    Your y'' matches the one I have above. However isn't sin(0)=0? So the C1 and C3 terms would go away when evaluated at 0? Leaving:

    y''(0)= C1(-2e^0(sin(0))+C2(2e^0(cos(0)))+C3(2e^0(sin(0)))
    +C4(-2e^0(cos(0)))
    =>0=0+2C2+0-2C4
    ?
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  12. #12
    Super Member Random Variable's Avatar
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    Quote Originally Posted by mmattson07 View Post
    Your y'' matches the one I have above. However isn't sin(0)=0? So the C1 and C3 terms would go away when evaluated at 0? Leaving:

    y''(0)= C1(-2e^0(sin(0))+C2(2e^0(cos(0)))+C3(2e^0(sin(0)))
    +C4(-2e^0(cos(0)))
    =>0=0+2C2+0-2C4
    ?
    Yes. You're correct. My mistake.
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  13. #13
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    Ok so I had the last sign on y''' wrong yours was correct though. So we have
    y(0)= 0 =C1+C3
    y'(0)= 1 =C1+C2-C3+C4
    y''(0)= 0 =2C2-2C4
    y'''(0)= 2 =-2C1+2C2+2C3+2C4

    How do you deduct the constants from these? The only thing I could see is that
    C2=C4
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  14. #14
    Super Member Random Variable's Avatar
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    From the first equation we also know that  C_{1}=-C_{3}

    Then from the second equation  2C_{1}+2C_{2} =1 (1)

    And from the fourth equation  -4C_{1}+4C_{2}=2 or  -2C_{1}+2C_{2}=1 (2)

    Add (1) and (2) to get  4C_{2} = 2

    so  C_{2} = \frac{1}{2} which means that  C_{4}=\frac{1}{2}

    and going back to (1),  2C_{1}+2\Big(\frac{1}{2}\Big) = 1

    so  C_{1} = 0 which means that  C_{3}=0
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  15. #15
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    Ahh, and you can add those because they're both equal to 1?

    And the solution to the IVP is <br />
y(x) =  (\frac{1}{2}) e^{x} \sin x +(\frac{1}{2})e^{-x} \sin x<br />
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