find the particular solution to the differential equation th

Solution

f \'(x) = -cosx + e^(2x)/2 + c
Substitute 0 for x and 1/2 for f \'(x) and solve for c.
1/2 = -cos0 + e^(2(0))/2 + c
1/2 = -1 + 1/2 + c
c = 1
f \'(x) = -cosx + e^(2x)/2 + 1
f(x) = -sinx + e^(2x)/4 + x + d
1/4 = -sin0 + e^(2(0))/4 + 0 + d
1/4 = 1/4 + d
d = 0
f(x) = -sinx + e^(2x)/4 + x