I actually realized this shortly after posting the problem, but wouldn't sinx*cosx end up with the solution being 1/2*ex*sin(2x)? That's the next part I got stuck on. I completed the problem using only 1+/-2i to get the quadratic in hopes for partial credit and got y''-2y'+3/2
Homework Statement
Given a set of fundamental solutions {ex*sinx*cosx, ex*cos(2x)}
Homework Equations
y''+p(x)y'+q(x)=0
det W(y1,y2) =Ce-∫p(x)dx
The Attempt at a Solution
I took the determinant of the matrix to get
e2x[cos(2x)cosxsinx-2sin(2x)sinxcosx-cos(2x)sinxcosx-...