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Stuck on what appears to be simple ODE

  1. Nov 9, 2011 #1
    I am completely stuck on where to go with the following ODE:

    (D^4 + 1)y = 0 where D=d/dx

    I know that trying y=e^(rt) is the obvious solution, however, when you solve this you get r^2 = +-i. At this point I am unsure of what to do becuase if I take the square root of "i" I am unsure of how I will represent it in a general solution. Any assistance would be greatly appreciated, thanks!
  2. jcsd
  3. Nov 9, 2011 #2


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    An equation like r^4+1=0 means that there are 4 solutions. Instead of putting r^2=+/-i, you can look at r^2=e^(pi/2)*i or e^(3pi/2)*i. Taking the square root, you simply get r=+/-e^(pi/4)*i or +/-e^(3pi/4)*i.

    The trick is to put it in the phase explicitly, so that you have an answer.
  4. Nov 9, 2011 #3


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    (D^4 + 1)=(D^2+sqrt(2)D+1)(D^2-sqrt(2)D+1)
    =(D^2+i)(D^2-i)=(D-√ i)(D+√ i)(D-√- i)(D+√- i)
    So write a solution in terms of complex exponentials or products of exponentials and trigonometric functions like (e^r x)cos(a x) according to preference.
  5. Nov 9, 2011 #4



    [tex]{r}^{2}=\pm i [/tex]

    [tex] r=\pm\frac{1\pm i}{\sqrt{2}}[/tex]

    [tex]y={C}_{1}\exp \left(\frac{1+i}{\sqrt{2}}\right)+{C}_{2}\exp \left(\frac{1-i}{\sqrt{2}}\right)+{C}_{3}\exp \left(-\frac{1+i}{\sqrt{2}}\right)+{C}_{4}\exp \left(\frac{i-1}{\sqrt{2}}\right)[/tex]

    We can solve for the specific solution for y if and only if we know y(0), y'(0), y''(0), y'''(0) and y''''(0)
  6. Nov 10, 2011 #5
    thanks everyone, I greatly appreciate the help!
  7. Nov 12, 2011 #6
    Just realised I made a LaTeX error. Should be

    [tex]y={C}_{1}\exp \left(\frac{1+i}{\sqrt{2}}x\right)+{C}_{2}\exp \left(\frac{1-i}{\sqrt{2}}x\right)+{C}_{3}\exp \left(-\frac{1+i}{\sqrt{2}}x\right)+{C}_{4}\exp \left(\frac{i-1}{\sqrt{2}}x\right)[/tex]

    and you may want to use euler's formula to put in some trigonometric functions there.
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