# Solve the initial value problem. {d^2 s} / {dt^2} = -16 sin (4 t - {3 pi} / 2), s' (0) = 13, s...

## Question:

Solve the initial value problem.

{eq}\displaystyle \dfrac {d^2 s} {dt^2} = -16 \sin \left (4 t - \dfrac {3 \pi} 2\right ),\ s' (0) = 13,\ s (0) = -2 {/eq}

## Initial Value Problems (IVP)

An initial value problem or IVP is a differential equation for which the initial conditions are specified. The initial conditions specified determine a specific solution to the differential equation. An example of an IVP is radioactive decay which follows an exponential decay process and the initial value equals the mass of the radioactive material.

In general, an IVP is specified as a differential equation: \begin{align*} \dfrac{d^n x}{dt^n} &= f(t) \end{align*} with initial conditions \begin{align*} x^{n-1}(T) = c_{n-1}, x^{n-2}(T) = c_{n-2}, \cdots, x'(T) = c_1, x(T) = c_0 \end{align*} where {eq}T, c_{n-1}, c_{n-2}, \cdots c_0 {/eq} are constants.

The IVP's differential equation is

$$\displaystyle \dfrac {d^2 s} {dt^2} = -16 \sin \left (4 t - \dfrac {3 \pi} 2\right )$$ with initial conditions {eq}s' (0) = 13,\ s (0) = -2 {/eq}.

This is a second-order IVP because the derivative on the LHS of the differential equation contains the second derivation of {eq}s {/eq} with {eq}t {/eq}.

Let {eq}y = s'(t) = \dfrac{ds}{dt} {/eq}.

Then {eq}y(0) = s'(0) = 13 {/eq}.

Substitute {eq}y = s'(t) {/eq} into the IVP which yields: \begin{align*} \dfrac{d^2 s}{dt^2} &= -16 \sin \left (4 t - \dfrac {3 \pi} 2\right ) \\ \Rightarrow \dfrac{d}{dt} {ds}{dt} &= -16 \sin \left (4 t - \dfrac {3 \pi} 2\right ) \\ \Rightarrow \dfrac{dy}{dt} &= -16 \sin \left (4 t - \dfrac {3 \pi} 2\right )\\ \\ \int dy &= \int -16 \sin \left (4 t - \dfrac {3 \pi} 2\right ) dt \\ \Rightarrow y(t) &= 4 \cos \left (4 t - \dfrac {3 \pi} 2\right ) + c_1 \end{align*}

The constant {eq}c_1 {/eq} can be determined by plugging in {eq}t = 0, y(0) = 13 {/eq}. \begin{align*} y(0) &= 4 \cos \left (4 \times 0 - \dfrac {3 \pi} 2\right ) + c_1 \\ \Rightarrow 13 &= 4 \cos \left (-\dfrac {3 \pi} 2\right ) + c_1 \\ \Rightarrow c_1 &=13 \end{align*}

So, the intermediate solution is {eq}y(t) = 4 \cos \left (4 t - \dfrac {3 \pi} 2\right ) + 13 {/eq}.

Now, since {eq}\dfrac{ds}{dt} = y(t) {/eq}, \begin{align*} \dfrac{ds}{dt} &= y(t) \\ \Rightarrow \dfrac{ds}{dt} &= 4 \cos \left (4 t - \dfrac {3 \pi} 2\right ) + 13 \\ \int ds &= \int \left(4 \cos \left (4 t - \dfrac {3 \pi} 2\right ) + 13\right) dt \\ \Rightarrow s(t) &= \sin \left (4 t - \dfrac {3 \pi} 2\right ) + 13t + c_0 \end{align*}

This is the general solution of the IVP. The specific solution can be found by finding {eq}c_0 {/eq}.

The constant {eq}c_0 {/eq} can be determined from the initial condition {eq}s(0) = -2 {/eq}. \begin{align*} s(0) &= \sin \left (4 \times 0 - \dfrac {3 \pi} 2\right ) + 13 \times 0 + c_0 \\ \Rightarrow -2 &= 1 + c_0 \\ \Rightarrow c_0 &= -3 \end{align*}

Plugging back {eq}c_0 = -3 {/eq} back into the differential equation's solution yields: \begin{align*} s(t) = \sin \left (4 t - \dfrac {3 \pi} 2\right ) + 13t -3 \end{align*}

Therefore the solution to the initial value problem is {eq}\color{Green}{s(t) = \sin \left (4 t - \dfrac {3 \pi} 2\right ) + 13t -3} {/eq}. 