Question

Solve the following differential equation with initial conditions:

y"=e^(-2t) + 10^(4t); y(0)=1, y'(0) = 0

Answer 1

well, we have

y' = -1/2 e^(-2t) + 1/ln10 * 1/4 10^(4t) + c
y'(0)=0 means -1/2 + 1/(4 ln10) + c = 0, so
c = 1/2 - 1/(4 ln10)

Now follow the same logic to get y(t)