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

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Answer 1 · 2016-08-26T08:29:29

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

# 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#