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Higher Order Differential Equation

  1. Jul 19, 2015 #1
    1. The problem statement, all variables and given/known data

    ##y^{(4)} + y = 0, y(0)=0, y'(0)=0,y''(0)=-1,y'''(0)=0##

    My issue with this equation is not with the steps, I don't believe but the solving of the IVP, the derivatives of my solution end up being close to 32 terms long, and I was wondering if there is any shorter method I could use without having to figure out the close to 50 terms for all 3 derivatives combined.



    2. Relevant equations


    3. The attempt at a solution

    upload_2015-7-19_22-58-46.png

    (ignore the ##c_3## next to my ##c_4## term, that was a typo)
     
  2. jcsd
  3. Jul 19, 2015 #2
  4. Jul 19, 2015 #3
  5. Jul 20, 2015 #4
    You can always plug it back in and show with pencil and paper that the solution satisfies the original diff eq and all the initial conditions.
     
    Last edited: Jul 20, 2015
  6. Jul 20, 2015 #5

    SteamKing

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    "pencul and panper"?
     
  7. Jul 20, 2015 #6
    Sorry about that. Typing too fast with too little coffee.
     
  8. Jul 20, 2015 #7

    HallsofIvy

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    Dr. Courtney is extremely old- "pencul and panper" were used before pencil and paper were invented.
     
  9. Jul 20, 2015 #8

    vela

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    Combine terms as you go along and look for patterns. For example, consider the terms ##f(x) = c_1 e^{ax}\cos ax + c_2 e^{ax}\sin ax##. First, factor the exponential out so you only have to do the product rule once. When you differentiate, you get
    \begin{align*}
    f'(x) &= a e^{ax}(c_1 \cos ax + c_2\sin ax) + e^{ax}(-a c_1 \sin ax + a c_2 \cos ax) \\
    &= a e^{ax} [(c_1+c_2)\cos ax + (c_2-c_1)\sin ax].
    \end{align*} Note that this is essentially the same form you started with, so you can easily write down the second derivative without much effort:
    \begin{align*}
    f''(x) &= a^2 e^{ax} [((c_1+c_2)+(c_2-c_1)) \cos ax + ((c_2-c_1)-(c_1+c_2))\sin ax] \\
    &= a^2 e^{ax} [2c_2 \cos ax - 2c_1 \sin ax] \\
    &= 2 a^2 e^{ax} [c_2 \cos ax - c_1 \sin ax]
    \end{align*} So in comparison to f(x), differentiating twice produced an overall factor of ##2a^2##, changed ##\sin x## into ##\cos x##, and changed ##\cos x## into ##{-\sin x}##. You should be able to convince yourself that ##f''''(x) = -4a^4 e^{ax}(c_1 \cos ax + c_2 \sin ax)##.
     
    Last edited: Jul 20, 2015
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