Antiderivatives: Find f if

Question

f''(x) = 3e^x - 2sinx, f(0)=3, f'(π/2)=0

Answer 1

y' = 3e^x + 2cosx + c

0 = 3e^(pi/2) + 2cos(pi/2) + c

0 = 3e^(pi/2) + c <--- cosine of 90 is zero

c = -3e^(pi/2)

y = 3e^x + 2 sinx - 3e^(pi/2) x + k

3 = 3 + k <--- sine of zero is zero, and zero kills 3e^(pi/2)

So

k = 0

y = 3e^x + 2 sinx - 3e^(pi/2) x