First, notice that \[ \DiffFrac {y}{x} = \frac {\Diff {y} / \Diff {\theta }}{\Diff {x} / \Diff {\theta }} = \frac {\Diff {r} / \Diff {\theta } \cdot \sin \theta + r \cdot \cos \theta }{\Diff {r} / \Diff {\theta } \cdot \cos \theta - r \cdot \sin \theta }. \]
Therefore, the original differential equation reduces to \[ (r \sin \theta + r \cos \theta ) \frac {\Diff {r} / \Diff {\theta } \cdot \sin \theta + r \cdot \cos \theta }{\Diff {r} / \Diff {\theta } \cdot \cos \theta - r \cdot \sin \theta } = r \sin \theta - r \cos \theta \] which further reduces to (since \(r \neq 0\)) \[ (\sin \theta + \cos \theta ) \left [\DiffFrac {r}{\theta } \cdot \sin \theta + r \cos \theta \right ] = (\sin \theta - \cos \theta ) \left [\DiffFrac {r}{\theta } \cdot \cos \theta - r \sin \theta \right ]. \]
Expanding the brackets and cancelling the equivalent terms gives us \[ r \cos ^2 \theta + \DiffFrac {r}{\theta }\sin ^2\theta = - \DiffFrac {r}{\theta } \cos ^2 \theta - r \sin ^2 \theta , \] which reduces to (due to the Pythagoras Theorem \(\sin ^2 \theta + \cos ^2 \theta = 1\)), \[ \DiffFrac {r}{\theta } + r = 0, \] as desired.
The rearrangement (since \(r \neq 0\)) \[ \frac {\Diff {r}}{r} = - \Diff {\theta } \] shows that the solution to this differential equation must satisfy that (since \(r > 0\)) \[ \ln r = - \theta + C, \] i.e. \[ r = A \exp (- \theta ), \] where \(A > 0\).
For critical values, notice that when \(\theta = 0\), \(r = A\), and when \(\theta = 2\pi \), \(r = \frac {A}{\exp 2\pi }\), and that \(r\) is decreasing with \(\theta \). The graph will look like a spiral
A sketch is shown below, for \(\theta \in [0, 2\pi )\).
Similar to the previous part, the equation reduces to \[ \left (\sin \theta + \cos \theta - \cos \theta \cdot r^2\right ) \left [\DiffFrac {r}{\theta } \cdot \sin \theta + r \cos \theta \right ] = \left (\sin \theta - \cos \theta - \sin \theta \cdot r^2\right ) \left [\DiffFrac {r}{\theta } \cdot \cos \theta - r \sin \theta \right ], \] and hence, by expanding brackets and eliminating terms, \[ \DiffFrac {r}{\theta } \sin ^2 \theta + r \cos ^2 \theta - r^3 \cos ^2 \theta = -r \sin ^2 \theta - \DiffFrac {r}{\theta } \cos ^2\theta + r^3 \sin ^2 \theta , \] which then simplifies to \[ \DiffFrac {r}{\theta } + r - r^3 = 0. \]
Notice that \(r = 1\) is a solution to this differential equation. Therefore, rearranging terms, we have \[ \frac {\Diff {r}}{r^3 - r} = \Diff {\theta }. \]
By partial fractions \[ \frac {1}{r^3 - r} = -\frac {1}{r} + \frac {1}{2(r + 1)} + \frac {1}{2(r - 1)}, \] we therefore must have \[ \left [-\frac {1}{r} + \frac {1}{2(r + 1)} + \frac {1}{2(r - 1)}\right ] \cdot \Diff {r} = \Diff \theta . \]
This therefore means that \[ \frac {1}{2} \ln \abs *{r + 1} + \frac {1}{2} \ln \abs *{r - 1} - \ln \abs *{r} = \theta + C, \] for some constant \(C \in \RR \).
Combining logarithms and absolute values gives us \[ \ln \abs *{\frac {r^2 - 1}{r^2}} = 2\theta + C, \] and therefore, \[ \frac {r^2 - 1}{r^2} = \pm \exp C \cdot \exp (2\theta ), \] and this can be simplified to \[ 1 - \frac {1}{r^2} = \pm \exp C \cdot \exp (2\theta ), \] and therefore \[ r^2 = \frac {1}{1 \mp \exp C \cdot \exp (2\theta )}. \]
Let \(A = \mp \exp C \neq 0\), and therefore \[ r^2 = \frac {1}{1 + A \exp (2 \theta )}. \]
Notice when \(r = 1\), \(r\) satisfies that \[ r^2 = 1, \] so the general solution will be \[ r^2 = \frac {1}{1 + A \exp (2 \theta )} \] for \(A \in \RR \) which this equation makes sense.
We restrict ourselves to \(\theta \in [0, 2\pi )\).
Notice that, this equation makes sense for all \(A \geq 0\), since the denominator is obviously non-negative.
For \(A < 0\), the denominator is decreasing in \(\theta \), and we would like it to be greater than zero for some \(\theta \in [0, 2\pi )\). Therefore, we would like the maximum possible value of the denominator to be greater than, that is when \(\theta = 0\): \[ 1 + A \exp 0 > 0, \] which gives \(A > -1\).
We consider three cases where \(r > 0\), i.e., \[ r = \frac {1}{\sqrt {1 + A \exp (2\theta )}}. \]
Notice this always passes through \(\left (\frac {1}{\sqrt {1 + A}}, 0\right )\).
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When \(-1 < A < 0\), the curve is not defined for \[ 1 + A \exp (2\theta ) \leq 0, \] and this is precisely when \[ \exp 2\theta \geq -\frac {1}{A}, \] which is \[ \theta \geq \frac {1}{2} \cdot \ln \left (-\frac {1}{A}\right ). \]
This means the curve will have an asymptote of line \[ \theta = \frac {1}{2} \cdot \ln \left (-\frac {1}{A}\right ). \]
Also note that \(r\) is increasing in \(\theta \) in this case, and \(r \to \infty \) as \(\theta \to \) the asymptote.
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When \(A = 0\), notice this just gives \(r = 1\), which is a circle with radius 1 centred at the origin.
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In the final case where \(A > 0\), the following case arises.
