I'm trying to use the variation of parameters method to find the geneneral solution of this equation:

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 cos(x) + f'2 sin(x) = 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 cos (x) + f'2 sin(x) = 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 = -2(tan^2(x)) * sin(x)

f'2 = 2(tan^2(x)) * cos(x)

But something is not right with this and I end up with a screwball answer. My yc is correct though. The answer in the back of the book is:

y=(e^-x)(c1cos(x) + c2sin(x) - 4 + 2sin(x)*ln(sec x + tan x))

I've worked through this 5 times and I keep getting the same answer. Any suggestions? Thanks a ton,

JN