# What is a solution to the differential equation tantheta(dr)/(d(theta))+r=sin^2theta where 0<theta<pi/2?

Aug 26, 2016

$r = \frac{1}{3} {\sin}^{2} \theta + C \csc \theta$

#### Explanation:

$\tan \theta \frac{\mathrm{dr}}{d \theta} + r = {\sin}^{2} \theta$

multiply across by $\cos \theta$
$\sin \theta \frac{\mathrm{dr}}{d \theta} + r \cos \theta = {\sin}^{2} \theta \cos \theta$

now look at the LHS closely

$\frac{\mathrm{dr}}{d \theta} \sin \theta + r \cos \theta = \frac{d}{d \theta} \left(r \setminus \sin \theta\right)$

so

$\frac{d}{d \theta} \left(r \setminus \sin \theta\right) = {\sin}^{2} \theta \cos \theta$

$\implies r \setminus \sin \theta = \int {\sin}^{2} \theta \cos \theta \setminus d \theta$

$\implies r \setminus \sin \theta = \frac{1}{3} {\sin}^{3} \theta + C$

$r = \frac{1}{3} {\sin}^{2} \theta + C \csc \theta$