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Thread: 2nd order differential complex!

  1. #1
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    2nd order differential complex!

    I need help with the first part mostly!

    The Question:

    $\displaystyle \frac { { d }^{ 2 }y }{ { d }t^{ 2 } } +y=\quad (2cosx)+\quad (2){ e }^{ x }$

    for the associated homogeneous

    D = d/dx

    $\displaystyle (D^2 + 1)y = 0 $

    roots a = b = i

    Trial solution:

    $\displaystyle y = (A +Bx)e^ix $

    when you sub it into the homogeneous, the real parts cancel but the imaginary doesn't!

    $\displaystyle -Ae^ix -Bxe^ix + 2Bie^ix +Ae^ix +Bxe^ix $

    I was wondering why this is?


    Am i right in thinking the trial solution is wrong?

    For the particular solutions of the other parts I got

    for $\displaystyle 2cosx $

    I got $\displaystyle { y }_{ 2 } = x^2cosx$

    for $\displaystyle 2{ e }^{ x }$

    I got $\displaystyle { e }^{ x }$

    please verify these!

    thanks
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  2. #2
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    Re: 2nd order differential complex!

    The CF will be $\displaystyle Acosx+Bsinx,$ your trial solution should be $\displaystyle y=x(Ccosx+Dsinx)+Ee^{x}.$
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  3. #3
    MHF Contributor
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    Re: 2nd order differential complex!

    When you have a complex conjugate pair (i and -i) of roots, then you use the cosine and sine functions. In your case, the solution to the homogeneous equation is $\displaystyle y=c_1cos{x}+c_2\sin{x}$.

    Your particular solution for $\displaystyle 2\cos{x}$ is incorrect. Try $\displaystyle x\sin{x}$.

    - Hollywood
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