# Homework Help: Method od undetermined coeff help needed.

1. May 5, 2007

### HappMatt

1. The problem statement, all variables and given/known data
Ive been working on this problem for way too long and i can get it almost right except im missing a factor of (1/12)e^4x and for the life of me i cant figure out why. Ive been mostly trying to use the methode of undeermined coefficients to no luck and so have also tried variation of parameters and still something is not going right. I have run the equation through my 89 and i have the answer i just cant seem to get it by hand.

2. Relevant equations
y''-2y'-8y=3*e^(4x)-5x^2
complimentary solution=yc=C1*e^(4x)+C28*e^(-2x)
as for finding the particular solution(yp) i think this is where my problem is.

the actual solution as given by my ti 89 is y=((x/2)
+C1-(1/12))*e^(4x)+C2*e^(-2x)+(5x^2)/8-(5x)/16+15/64

3. The attempt at a solution
using yp=Axe^(4x) +Bx^2+Cx+d i got it all right except that im missing the (1/12)*e^4x value and ive tried to many varriations to list them all here but i have litterally spent hours on this and i probally should have posted this back when i was still at only 3 hours worth of time into it but im well beyond that now and its time for a little rest so i can wake up and keep hammering at this one till i get it.

2. May 5, 2007

### Pseudo Statistic

When dealing with undetermined co-efficients, you'd probably find it best to, rather than tackle both right hand terms (3e^(4x) - 5x^2) at the same time, deal with them as two seperate problems; that is, guess their particular solution for:
y'' - 2y' - 8y = 3e^(4x)
Then do the same for y'' - 2y' - 8y = -5x^2 and add up your solutions.
However, it seems like that's not the problem you're having; when you say you're missing this "1/12 e^4x", that doesn't quite add up, since 1/12 e^4x contributes to the homogeneous solution.
That is, unless you're given some initial conditions you're not telling us about? ;)

3. May 5, 2007

### HappMatt

thanks for the advice Psuedo as far as seperating the particular solution Ive tried that and recived the same results. As for the (1/12)*e^(4x) Im mostly sure that is part of the particular solution since the homogeneous solution is just C2e^4x +C1e^(-2x)

4. May 5, 2007

### HappMatt

tried this probleb again for the 10000th and i finally got the answer using variation of parameters.