Question

Last problem of the day!
Find f.
f ''(t) = 2e^t + 3 sin t, f(0) = 0, f(π) = 0

Answer 1

f ''(t) = 2e^t + 3 sin t
f'(t) = ∫f ''(t)dt
= ∫ 2e^t + 3 sin t dt
= 2e^t - 3 cos t + C1
f(t) = ∫f '(t)dt
= ∫(2e^t - 3 cos t + C1)dt
= 2e^t - 3 sin t + (C1)t +C2

f(0) = 2e^(0) - 3 sin (0) + (C1)(0) +C2 = 0
2 -0 +0 +C2 =0
C2 = -2

f(π) = 0
2e^(π) - 3 sin (π) + (C1)(π) +C2 = 0
2e^(π) -0 + (C1)(π) -2 =0
C1 = [2-2e^(π)] /π

therefore
f(t)= 2e^t - 3 sin t + { [2-2e^(π)] t} /π -2