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2nd order O.D.E

  1. Nov 13, 2005 #1
    for the following question:

    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


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    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? @@
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