Question: Identify each of the differential equations in Problems 1to 24 as to type (for example, separable, linear first order, linear second order, etc.), and then solve it.

$\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathit{d}\mathit{r}\mathbf{-}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathit{d}\mathit{\theta }\mathbf{=}\mathit{r}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathit{d}\mathit{\theta }$

The differential equation is $\sin \theta \cos \theta dr-{\sin }^2\theta d\theta =r{\cos }^2\theta d\theta$.

When fand its derivatives are inserted into the equation, a solution is a function y = f(x) that solves the differential equation. The highest order of any derivative of the unknown function appearing in the equation is the order of a differential equation.

A differential equation of the form$\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{a}\mathbf{\ne }\mathit{b}$ has general solution$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{a}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{b}\mathbf{x}}\mathbf{.}$

Consider the equation.

$\sin \theta \cos \theta dr-{\sin }^2\theta d\theta =r{\cos }^2\theta d\theta$

Rearrange above equation.

$\begin{array}{c}\frac{\sin \theta \cos \theta dr}{\sin \theta \cos \theta d\theta }-\frac{{\sin }^2\theta d\theta }{\sin \theta \cos \theta d\theta }=\frac{r{\cos }^2\theta d\theta }{\sin \theta \cos \theta d\theta }\\ \frac{dr}{d\theta }-\frac{\sin \theta }{\cos \theta }=\frac{r\cos \theta }{\sin \theta }\\ \frac{dr}{d\theta }-\frac{r\cos \theta }{\sin \theta }=\frac{\sin \theta }{\cos \theta }\end{array}$

The above equation is in the form of$y^{\text{'}}+Py=Q$where $P=-\frac{\cos \theta }{\sin \theta },Q=\frac{\sin \theta }{\cos \theta }$.

The integrating factor is,

$\begin{array}{c}I=\int Pd\theta \\ =\int --\frac{\cos \theta }{\sin \theta }d\theta \\ =-\mathrm{lnsin}\theta \\ =-\ln {\left(\sin \theta \right)}^{-1}\end{array}$

The solution of differential equation is,

$\begin{array}{c}r{e}^I=\int Q{e}^{I}d\theta r{e}^{\ln {\left(\sin \theta \right)}^{-1}}\\ =\int \frac{\sin \theta }{\cos \theta }{e}^{\ln {\left(\sin \theta \right)}^{-1}}d\theta r\frac{1}{\sin \theta }\\ =\int \frac{\sin \theta }{\cos \theta }\frac{1}{\sin \theta }d\theta r\frac{1}{\sin \theta }\\ =\int \sec \theta d\theta \end{array}$

Further solve,

$r=\sin \theta [\ln \left(\sec \theta +\tan \theta \right)+C]$

Thus, the solution of given differential equation is $r=\sin \theta [\ln \left(\sec \theta +\tan \theta \right)+C]$.