Question

In problems 1–24 Find the general solution of the given differential equation. Give the largest interval I over which the general solution is defined. Determine whether there are any transient terms in the general solution.

$\frac{dr}{d\theta }+r\sec \theta =\cos \theta$

Short answer

The general solution and interval is$r\left(\sec \theta +\tan \theta \right)=\theta -\cos \theta +C,\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

Step-by-step solution

Concept of Linear Differential Equation

$\frac{dy}{dx}+py=q$where p and q can be constants or variables.

We will find integrating factor to find the general solution

Finding an integrating factor

$e^{\int \sec \theta d\theta }=e^{\ln (\sec \theta +\tan \theta )}=\sec \theta +\tan \theta$

Multiplying integrating factor to both sides

$$\begin{gathered} (\sec \theta +\tan \theta )\left[\frac{dr}{d\theta }+r\sec \theta \right]=(\sec \theta +\tan \theta )[\cos \theta ] \\ \frac{d}{d\theta }\left[\right(\sec \theta +\tan \theta )r]=(1+\sin \theta ) \end{gathered}$$

Integrating both sides

$$\begin{gathered} \int d\left[\right(\sec \theta +\tan \theta )r]=\int (1+\sin \theta )d\theta \\ (\sec \theta +\tan \theta )r=\theta -\cos \theta +C \\ r=\frac{\theta -\cos \theta +C}{\sec \theta +\tan \theta } \\ r\left(\sec \theta +\tan \theta \right)=\theta -\cos \theta +C \end{gathered}$$

Finding an Interval

The solution is defined at $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

Hence, the solution is $r\left(\sec \theta +\tan \theta \right)=\theta -\cos \theta +C,\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$