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) +...