Question

In Exercises \(11-20,\) solve the initial value problem explicitly. \(\frac{d v}{d t}=4 \sec t \tan t+e^{t}+6 t\) and \(v=5\) when \(t=0\)

Short answer

Therefore, the solution to the initial value problem is \(v(t)=4\sec t + e^{t} + 3t^{2}\).

Step-by-step solution

Integration

Integrate the first order non-homogeneous ordinary differential equation by isolating the \(\frac{dv}{dt}\) on one side and integrating the other side with respect to \(t\) to get \(v(t)\). \nThus, \(\int dv = \int (4\sec t\tan t+e^t +6t)dt)\) leading to \(v(t) = 4\int(\sec t\tan t) dt + \int(e^{t})dt + \int(6t)dt + C\), where \(C\) is the constant of integration.

Evaluate Integrals

Evaluate each integral separately. All of these are standard integrals: \(\int\sec t\tan t dt=\sec t\), \(\int e^t dt=e^t\) and \(\int 6t dt=3t^2\). Thus, \(v(t)=4\sec t + e^{t} + 3t^{2} + C\).

Apply Initial Condition

Use the initial condition \(v=5\) when \(t=0\) to solve for the constant \(C\). Substituting these values into the equation we get: \(5 = 4(1) + 1 + 0 + C\). Solving for \(C\) yields \(C = 0\).

Final Solution

Substitute \(C=0\) back into the equation from step 2 to get the final solution to the initial value problem, which is \(v(t)=4\sec t + e^{t} + 3t^{2}\).