2nd order O.D.E

1. Nov 13, 2005

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?

2. Nov 13, 2005

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.

Last edited: Nov 13, 2005
3. Nov 13, 2005

HallsofIvy

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 by a moderator: Nov 13, 2005
4. Nov 13, 2005

asdf1

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