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ckp
Sep10-09, 07:04 PM
dr/d(theta) + r*sec(theta) = cos(theta)

apparently the solution is (sec(theta) + tan(theta))*r = theta - cos(theta) + c

but i have no idea how to get there.

i am using a technique from the book, but it yeilds an answer with lots of e's and an integral i cant solve.

this is my work. hope it doesnt look like a complete mess

ce^(-(sec(theta)tan(theta))) + e^(-(sec(theta)tan(theta)))(integral([e^(sec(theta)tan(theta))]cos(theta))d(theta)]

and i have no idea what to do from here.

rock.freak667
Sep10-09, 07:10 PM
\frac{dr}{d \theta} + r sec\theta=cos\theta


looks a bit like

\frac{dy}{dx} + yP(x)=Q(x)

doesn't it? So try an integrating factor.

ckp
Sep10-09, 07:15 PM
i was trying to use an integrating factor, but im not sure if im using it right. the long mess is what i end up with and i cant go any further. my book doesnt do a very good job of explaining it

ckp
Sep10-09, 07:30 PM
ha, nevermind. I'm sorry, I messed up the integral of sec(X) >_>

got it now.