differential equation with trig

**genlovesmusic09** · December 16th 2009, 08:18 AM

$\tan (\theta )\frac{{dr}} {{d\theta }} + r = {\sin ^2}(\theta ),\frac{\pi } {4} < \theta < \frac{\pi } {2} $

not sure where to begin...

**tonio** · December 16th 2009, 08:40 AM

Originally Posted by genlovesmusic09

$\tan (\theta )\frac{{dr}} {{d\theta }} + r = {\sin ^2}(\theta ),\frac{\pi } {4} < \theta < \frac{\pi } {2} $

not sure where to begin...

$\frac{dr}{d\theta}=\frac{\sin^2\theta-r}{\tan\theta}$ $=\sin\theta\cos\theta-r\frac{\cos\theta}{\sin\theta}\Longrightarrow r\theta=\int \sin\theta\cos\theta\,d\theta-r\int \frac{\cos\theta}{\sin\theta}\,d\theta$ ...now end the exercise.

Tonio

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