y" + 2y' + 2y = 2(e^-x)(tan^2(x))
Here's the work I've done:
1. r^2 + 2r + 2 =0, so by quadratic, r= -1 + i, r=-1 - i
2. yc = c1(e^-x)cos(x) + c2(e^-x)sin(x)
3. yp = f1(e^-x)cos(x) + f2(e^-x)sin(x)
4. f'1(e^-x)cos(x) + f'2(e^-x)sin(x) = 0; f'1 + f'2 = 0
5. f'1(-(e^-x)sin(x)-(e^-x)cos(x)) + f'2((e^-x)cos(x)-(e^-x)sin(x)) =
(e^-x)2(tan^2(x)
The system of equations simplifies to:
f'1 + f'2 = 0
f'1(-sin(x)-cos(x)) + f'2(cos(x) + sin(x)) = 2(tan^2(x))
Then I used Cramer's Rule to solve for f'1 and f'2 which is:
f'1 = -(tan^2(x))/(cos(x))
f'2 = (tan^2(x))/(cos(x))
But something is not right with this and I end up with an answer that is grossly different than that in the back of the book. I've worked through this 3 times and I keep getting the same answer. Any suggestions? Thanks a ton,
JN