Question

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

Answer 1

`r = 1/3 sin^2 theta + C csc theta`

Explanation:

`tantheta(dr)/(d theta)+r=sin^2theta`

multiply across by `cos theta`
`sin theta(dr)/(d theta)+ r cos theta =sin^2theta cos theta`

now look at the LHS closely

`(dr)/(d theta) sin theta + r cos theta = d/(d theta) ( r \ sin theta )`

so

`d/(d theta) ( r \ sin theta )=sin^2theta cos theta`

`implies r \ sin theta = int sin^2theta cos theta \ d theta`

`implies r \ sin theta = 1/3 sin^3 theta + C`

`r = 1/3 sin^2 theta + C csc theta`