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4th order equation

  1. Apr 11, 2007 #1
    I have a question regarding a 4th order differential equation from an exam i just took.

    we were asked to solve y^(4)-1=5 given y'(0)=y''(0)=y^(3)(0)=0

    I started by factoring down to (r-1)(r+1)(r^2+1)=5.

    I then found my general solution to be y=C_1e^6x+C_2e^4x+C_3e^2x+C_4e^-2x

    Obviously I would then be left with four equations with four unknowns to solve for my constants. Would I need to use a solver and/or hand solve the equations in order to find the constants? I didn't have time to do that so I just put the equations into a matrix and said that the constants were all equal to zero.

    Just thought I would ask what the right approach would be (a little to anxious to wait another week)
  2. jcsd
  3. Apr 11, 2007 #2
    Can you clarify what your DE is:

    Is it [itex]y^{(4)} - y = 5[/itex], or [itex]y^{(4)} - 1 =5[/itex]?

    In the second case, rearrange it to [itex]y^{(4)} = 6.[/itex] The answer is just some 4th order polynomial (the coeffs are easy so I'll let you work them out. Think about it for a minute :smile:).

    In the first case the answer is even simpler; y(x) = -5. :wink:

    Edit: Ignore this and see below.
    Last edited: Apr 11, 2007
  4. Apr 11, 2007 #3
    If what you wrote is correct, then the solution is [itex]y(x)=\frac{x^4}{4}+c[/itex], where [itex]c[/itex] is a constant.

    If what you meant is [itex]y^{(4)}-y=5[/itex], then you are missing one condition on [itex]y(0)[/itex], and your solution is incorrect.

    What went wrong you may ask. Well, the solution of the second equation is the sum of the homogeneous equation [itex]y_h^{(4)}-y_h=0[/itex], plus the solution of the particular equation [itex]y_p^{(4)}-y_p=5[/itex]. The particular solution is clearly [itex]y_p=-5[/itex], so you only have to solve the homogeneous equation, which indeed has the form


    where [itex]r_i[/itex] are the roots of the characteristic equation [itex]r^4-1=0[/itex]. Therefore, the solution is


    and evaluating on the initial conditions,

    [tex]y(x)=\frac{y_0+5}{2}\left[\cosh x+\cos x\right]-5,[/tex]

    where [itex]y(0)=y_0[/itex].
    Last edited: Apr 11, 2007
  5. Apr 11, 2007 #4
    Whoops, you're right. I forgot about the missing fourth init. cond.!
    Last edited: Apr 11, 2007
  6. Apr 11, 2007 #5
    Lets hope the missing condition is [itex]y(0)=-5[/itex] ;)
  7. Apr 11, 2007 #6
    Indeed! I was about to post the problem with regard to the 2 in front of the cos term (I needed to check that I wasn't making another dumb mistake, first!), but I see that you've already gotten to it. :biggrin:
  8. Apr 11, 2007 #7
    Lol, I always have to edit my posts like 7 times :P
  9. Apr 11, 2007 #8
    Yes, it seems like a common phenomenon with mine as well :wink:.
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