1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: 4th order equation
  1. 4th order DE (Replies: 4)

  2. 4th order DE (Replies: 1)