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!

2nd order O.D.E

  1. Nov 13, 2005 #1
    for the following question:
    y``+8y`+16y=64cosh4x

    my problem:
    suppose yp=c1cosh4x+c2sinh4x
    then yp`=4c1sinh4x+4c2cosh4x
    so yp``=16c1cosh4x+16sinh4x

    so 16c1cosh4x+16sinh4x +8(4c1sinh4x+4c2cosh4x)+16(c1cosh4x+c2sinh4x)= (32c1+32c2)cosh4x+(32c2+32c1)sinh4x

    which implies that (32c1+32c2)=0 and (32c2+32c1)=0 which is paradoxing!!!
    does anybody know what went wrong?
     
  2. jcsd
  3. Nov 13, 2005 #2
    Try reduction of order. In using this method you can take just part of the solution to the associated homogeneous equation. I would try [tex]y = u\left( x \right)e^{ - 4x} [/tex]. Any 'non-exponentials' eg polynomials in the complimentary solution get absorbed into u(x). Try the substitution I suggested and see if it leads anywhere.

    Edit: The method of undetermined coefficients only works for a few types of functions.

    Edit 2: I made an error in my suggested substitution. Fixed now.
     
    Last edited: Nov 13, 2005
  4. Nov 13, 2005 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Did you notice that e-4x and xe-4x are solutions to the homogeneous equation? Since 64 cosh 4x= 32(e4x+ e-4x) , you will have to multiply by x2. I would recommend trying y= Ae4x+ Bx2e-4x.
     
    Last edited: Nov 13, 2005
  5. Nov 13, 2005 #4
    hmmm... then what's wrong with my orignal assumptions? @@
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: 2nd order O.D.E
  1. 2nd order O.D.E. (Replies: 5)

  2. Second order O.D.E (Replies: 1)

Loading...