91Ó°ÊÓ

The differential equation is $\sin \mathrm{θ}\cos \mathrm{θ}dr−{\sin }^2\mathrm{θ}d\mathrm{θ}=r{\cos }^2\mathrm{θ}d\mathrm{θ}$.

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$\mathbf{(}\mathit{D}\mathbf{−}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{−}\mathit{b}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\text{ }}\mathit{a}\mathbf{≠}\mathit{b}$ 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.

$\sin \mathrm{θ}\cos \mathrm{θ}dr−{\sin }^2\mathrm{θ}d\mathrm{θ}=r{\cos }^2\mathrm{θ}d\mathrm{θ}$

Rearrange above equation.

$\begin{array}{c}\frac{\sin \mathrm{θ}\cos \mathrm{θ}dr}{\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{θ}}−\frac{{\sin }^2\mathrm{θ}d\mathrm{θ}}{\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{θ}}=\frac{r{\cos }^2\mathrm{θ}d\mathrm{θ}}{\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{θ}}\\ \frac{dr}{d\mathrm{θ}}−\frac{\sin \mathrm{θ}}{\cos \mathrm{θ}}=\frac{r\cos \mathrm{θ}}{\sin \mathrm{θ}}\\ \frac{dr}{d\mathrm{θ}}−\frac{r\cos \mathrm{θ}}{\sin \mathrm{θ}}=\frac{\sin \mathrm{θ}}{\cos \mathrm{θ}}\end{array}$

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

The integrating factor is,

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

The solution of differential equation is,

$\begin{array}{c}r{e}^I=∫Q{e}^{I}d\mathrm{θ}r{e}^{\ln {\left(\sin \mathrm{θ}\right)}^{−1}}\\ =∫\frac{\sin \mathrm{θ}}{\cos \mathrm{θ}}{e}^{\ln {\left(\sin \mathrm{θ}\right)}^{−1}}d\mathrm{θ}r\frac{1}{\sin \mathrm{θ}}\\ =∫\frac{\sin \mathrm{θ}}{\cos \mathrm{θ}}\frac{1}{\sin \mathrm{θ}}d\mathrm{θ}r\frac{1}{\sin \mathrm{θ}}\\ =∫\sec \mathrm{θ}d\mathrm{θ}\end{array}$

Further solve,

$r=\sin \mathrm{θ}[\ln \left(\sec \mathrm{θ}+\tan \mathrm{θ}\right)+C]$

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