Find f. f ''(t) = 3e^t + 8 sin(t), f(0) = 0, f(𝜋) = 0

Answer 1

To get f(t) from f"(t) we need to integrate f"(t) twice.

That will introduce two unknown constant a and b.

Then we calculate a and b from the boundary conditions for f(0) and f(π).

f"(t) = 3e^t + 8sin(t)

f'(t) = 3e^t - 8cos(t) + a

f(t) = 3e^t - 8sin(t) + at + b

Calculate the constant b from f(0) = 0:

3e⁰ - 8sin(0) + a x 0 + b = 0

b = -3

Calculate a from f(π) = 0:

3e^π - 8sin(π) + a x π - 3 = 0

a = (3 - 3e^π)/π

Finally

f(t) = 3e^t - 8sin(t) + (3 - 3e^π)t/π - 3