solve the initial value problem. dv/dt = 2/7 sec t tan t, v(π) = 2 v(t) =

Explanation:

Step1: Integrate the right - hand side

We know that the antiderivative of $\sec t\tan t$ is $\sec t$. So, integrating $\frac{2}{7}\sec t\tan t$ with respect to $t$ gives $v(t)=\frac{2}{7}\sec t + C$.

Step2: Use the initial condition

Substitute $t = \pi$ and $v(\pi)=2$ into $v(t)=\frac{2}{7}\sec t + C$. Since $\sec\pi=- 1$, we have $2=\frac{2}{7}\times(-1)+C$.

Step3: Solve for $C$

First, simplify the right - hand side of the equation $2=\frac{-2}{7}+C$. Then, $C = 2+\frac{2}{7}=\frac{14 + 2}{7}=\frac{16}{7}$.

Answer:

$v(t)=\frac{2}{7}\sec t+\frac{16}{7}$