Question

Solve the following initial value problem.

Answer 1

For starters,

$\cos(6t+\pi) = \cos(6t) \cos(\pi) - \sin(6t) \sin(\pi) = -\cos(6t)$

Now by the fundamental theorem of calculus, integrating both sides gives

$\displaystyle \frac{ds}{dt} = s'(0) + \int_0^t 36 \cos(6u) \, du = 100 + 6 \sin(6t)$

Integrating again, we get

$\displaystyle s(t) = s(0) + \int_0^t (100 + 6\sin(6u)) \, du = \boxed{100t - \cos(6t) + 1}$

Alternatively, you can work with antiderivatives, then find the particular constants of integration later using the initial values.

$\displaystyle \int \frac{d^2s}{dt^2} \, dt = \int 36\cos(6t) \, dt \implies \frac{ds}{dt} = 6\sin(6t) + C_1$

$\displaystyle \int \frac{ds}{dt} \, dt = \int (6\sin(6t) + C_1) \, dt \implies s(t) = -\cos(6t) + C_1t + C_2$

Now,

$s(0) = 0 \implies 0 = -1 + C_2 \implies C_2 = 1$

and

$s'(0) = 100 \implies 100 = 0 + C_1 \implies C_1 = 100$

Then the particular solution to the IVP is

$s(t) = -\cos(6t) + 100t + 1$

just as before.