Paradox in Solving y``+8y`+16y=64cosh4x

[asdf1]

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?

 

[Benny]

Try reduction of order. In using this method you can take just part of the solution to the associated homogeneous equation. I would try $$y = u\left( x \right)e^{ - 4x}$$. 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.

 

[HallsofIvy]

Did you notice that e^-4x and xe^-4x are solutions to the homogeneous equation? Since 64 cosh 4x= 32(e^4x+ e^-4x) , you will have to multiply by x². I would recommend trying y= Ae^4x+ Bx²e^-4x.

 

[asdf1]

hmmm... then what's wrong with my orignal assumptions? @@