Solve the initial value problem.
-$\frac { \mathrm { d } ^ { 2 } \mathrm {~s} } { \mathrm { dt } ^ { 2 } } = - 81 \sin \left( 9 \mathrm { t } - \frac { 3 \pi } { 2 } \right) , \mathrm { s } ^ { \prime } ( 0 ) = 7 , \mathrm {~s} ( 0 ) = - 2$
A) $s = \sin \left( 9 t - \frac { 3 \pi } { 2 } \right) - 3$
B) $s = \sin \left( 9 t - \frac { 3 \pi } { 2 } \right) + 6 t - 2$
C) $s = 9 \cos \left( 9 t - \frac { 3 \pi } { 2 } \right) + 7$
D) $s = \sin \left( 9 t - \frac { 3 \pi } { 2 } \right) + 7 t - 3$

Question

Quizplus · Accepted Answer

We can use integration twice to find the function s. First, integrating both sides with respect to t:
$$\mathrm{s}'(t) = \int \frac{\mathrm{d}^2 \mathrm{s}}{\mathrm{d}t^2} \mathrm{d}t = \int -81 \sin \left(9t - \frac{3\pi}{2}\right) \mathrm{d}t = 9\cos \left(9t - \frac{3\pi}{2}\right) + C_1$$
Using the initial condition $\mathrm{s}'(0) = 7$, we have:
$$7 = 9\cos \left(-\frac{3\pi}{2}\right) + C_1$$
$$C_1 = 7+9$$
$$C_1 = 16$$
Integrating again with respect to t:
$$\mathrm{s}(t) = \int \mathrm{s}'(t) \mathrm{d}t = \int \left(9\cos \left(9t - \frac{3\pi}{2}\right) + 16\right) \mathrm{d}t$$
$$\mathrm{s}(t) = \frac{\sin \left(9t - \frac{3\pi}{2}\right)}{9} + 16t + C_2$$
Using the initial condition $\mathrm{s}(0) = -2$, we have:
$$-2 = \frac{\sin \left(-\frac{3\pi}{2}\right)}{9} + C_2$$
$$C_2 = -2 + \frac{1}{9}$$
$$C_2 = -\frac{17}{9}$$
Therefore, the solution to the initial value problem is:
$$\mathrm{s}(t) = \frac{\sin \left(9t - \frac{3\pi}{2}\right)}{9} + 16t - \frac{17}{9}$$
Which matches choice D.

Solve the Initial Value Problem A) s=sin⁡(9t−3π2)−3s = \sin \left( 9 t - \frac { 3 \pi } { 2 } \right) - 3s=sin(9t−23π​)−3

Solve the Initial Value Problem A) $s = \sin \left( 9 t - \frac { 3 \pi } { 2 } \right) - 3$