Solve the initial value problem.
-$$\frac { \mathrm { dr } } { \mathrm { dt } } = 6 \mathrm { t } + \sec ^ { 2 } \mathrm { t } , \mathrm { r } ( - \pi ) = 5$$
A) $r = 3 t ^ { 2 } + \cot t + 5 - 3 \pi ^ { 2 }$
B) $r = 6 + \tan t - 1$
C) $r = 3 t ^ { 2 } + \tan t + 5 - 3 \pi ^ { 2 }$
D) $r = 6 t ^ { 2 } + \tan t + 5 - 6 \pi ^ { 2 }$

Question

Quizplus · Accepted Answer

We begin by integrating both sides of the given equation with respect to $t$:
$$r = 3t^2 + 	an t + C,$$
where $C$ is the constant of integration. To determine the value of $C$, we use the initial condition $r(-\pi)=5$. Substituting $t=-\pi$ and $r=5$, we get
$$5 = 3(-\pi)^2 + 	an(-\pi) + C = 3\pi^2 + C -\infty.$$
Since $	an(-\pi) = 	an(\pi) = 0$ and $\lim_{tightarrow -\pi^+} 	an t = -\infty$, we must have $C=5-3\pi^2$.

Therefore, the solution to the initial value problem is
$$r = 3t^2 + 	an t + 5 - 3\pi^2.$$

Solve the Initial Value Problem drdt=6t+sec⁡2t,r(−π)=5\frac { \mathrm { dr } } { \mathrm { dt } } = 6 \mathrm { t } + \sec ^ { 2 } \mathrm { t } , \mathrm { r } ( - \pi ) = 5dtdr​=6t+sec2t,r(−π)=5

Solve the Initial Value Problem $\frac { \mathrm { dr } } { \mathrm { dt } } = 6 \mathrm { t } + \sec ^ { 2 } \mathrm { t } , \mathrm { r } ( - \pi ) = 5$