Question

The following differential equation governs the evolution of variable $x(t)$ with time $t, t \geq 0$.
$\frac {d^2x}{dt^2} + 4x = e^{-t}$
Given the initial conditions $x = 0$ and $\frac {dx}{dt} = 0$ at $t = 0$, the value of $x$ at $t =\frac {\pi}{8}$ is _____ . (Rounded off to 3 decimal places)

Answer 1

The answer is: 0.064 or 0.065 (approx.)