Problem source

The curve $C$ has the differential equation in polar coordinates
\[
\frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r=5\sin3\theta,\qquad\text{for }\quad\frac{\pi}{5}\leqslant\theta\leqslant\frac{3\pi}{5},
\]
and, when $\theta=\dfrac{\pi}{2},$ $r=1$ and $\dfrac{\mathrm{d}r}{\mathrm{d}\theta}=-2.$ 
Show that $C$ forms a closed loop and that the area of the region enclosed by $C$ is 
\[
\frac{\pi}{5}+\frac{25}{48}\left[\sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right)\right].
\]

Solution source

First we seek the complementary function.
\begin{align*}
&& \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 0 \\
\Rightarrow && r &= A \sin 2\theta + B \cos 2 \theta
\end{align*}

Next we seek a particular integral, of the form $r = C \sin 3 \theta$.

\begin{align*}
&& \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 5 \sin 3 \theta \\
\Rightarrow && -9C \sin 3 \theta + 4C \sin 3 \theta &= 5 \sin 3 \theta \\
\Rightarrow && C &= -1 \\
\end{align*}

So our general solution is $A \sin 2\theta + B \cos 2 \theta -\sin 3 \theta$.

Plugging in boundary conditions we obtain:

\begin{align*}
\theta = \frac{\pi}{2}, r = 1: &&1 &=  -B +1 \\
\Rightarrow && B &= 0 \\
\theta = \frac{\pi}{2}, \frac{\d r}{\d \theta} = -2: && -2 &= -2A \\
\Rightarrow && A &= 1
\end{align*}

So the general solution is $r = \sin 2 \theta  - \sin 3 \theta = 2 \sin \left ( \frac{-\theta}{2} \right) \cos \left (\frac{5 \theta}{2} \right)$

First notice that for $\theta \in \left [\frac{\pi}{5}, \frac{3 \pi}{5} \right]$ this is positive, and it is zero on the end points, therefore we are tracing out a a loop.

The area of the loop will be:

\begin{align*}
A &= \int_{\pi/5}^{3\pi/5} \frac12 \left ( \sin 2 \theta - \sin 3 \theta \right)^2 \d \theta \\
&= \frac12\int_{\pi/5}^{3\pi/5}  \sin^2 2\theta + \sin^2 3 \theta - 2 \sin 2 \theta \cos 3 \theta  \d \theta \\
&= \frac12\int_{\pi/5}^{3\pi/5}  \frac{1-2 \cos 4 \theta}{2} + \frac{1-2 \cos6 \theta}{2} - \sin5 \theta-\cos\theta  \d \theta \\
&= \frac12 \left [\theta - \frac14 \sin 4 \theta-\frac16 \sin 6 \theta + \frac15 \cos 5 \theta  - \sin \theta \right]_{\pi/5}^{3\pi/5} \\
&= \frac{\pi}{5} +\frac{25}{48}\left [ \sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right) \right]
\end{align*}