Taking the derivative of both sides of the equation, we get f'(t) + f(t) = 0.
So, f'(t) = -f(t)
Letting y = f(t), we have dy/dt = -y
Separating variables, (1/y)dy = -dt
Integrate to get lnlyl = -t + C
lyl = e-t+C = eCe-t
y = Ce-t
From the original equation, if t = 0, we see that f(0) = 1.
So, Ce0 = 1. Therefore, C = 1
Thus, f(t) = e-t