Solve the initial value problem.
-$$\frac { \mathrm { ds } } { \mathrm { dt } } = ( 6 \mathrm { t } + 3 ) \sin \left( 3 \mathrm { t } ^ { 2 } + 3 \mathrm { t } \right) , \quad \mathrm { s } ( 0 ) = - 4$$
A) $s = \sin \left( 3 t ^ { 2 } + 3 t \right) - 4$
B) $s = \cos \left( 3 t ^ { 2 } + 3 t \right) - 5$
C) $s = - \cos \left( 3 t ^ { 2 } + 3 t \right) - 3$
D) $s = - \cos ( 6 t + 3 ) - \frac { 9 } { 4 }$

Question

Quizplus · Accepted Answer

To solve the initial value problem, we integrate the given differential equation $\frac { \mathrm { ds } } { \mathrm { dt } } = ( 6 \mathrm { t } + 3 ) \sin \left( 3 \mathrm { t } ^ { 2 } + 3 \mathrm { t } \right)$. The integral of the right side with respect to $t$ is $-\cos(3t^2 + 3t) + C$, since the derivative of $3t^2 + 3t$ is $6t + 3$, which matches the coefficient of the sine function, making it a straightforward application of the chain rule in reverse. Using the initial condition $s(0) = -4$, we find $C$ by substituting $t = 0$ into the integrated function, resulting in $-\cos(0) + C = -4$, which simplifies to $C = -3$ since $\cos(0) = 1$. Therefore, the solution is $s = - \cos \left( 3 t ^ { 2 } + 3 t \right) - 3$.

Solve the Initial Value Problem A) s=sin⁡(3t2+3t)−4s = \sin \left( 3 t ^ { 2 } + 3 t \right) - 4s=sin(3t2+3t)−4

Solve the Initial Value Problem A) $s = \sin \left( 3 t ^ { 2 } + 3 t \right) - 4$