Question

For the following differential equation, find the general solution:
$dy over dx equals fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction$

Solution

The given differential equation is

dy over dx space equals space fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction

dy space equals space fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction dx space space space or space space space space integral dy space equals space fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction dx

therefore space space space space space integral dy space equals space integral fraction numerator 2 space sin squared begin display style straight x over 2 end style over denominator 2 space cos squared space begin display style straight x over 2 end style end fraction dx space space or space space space integral space 1. space dy space equals space integral tan squared straight x over 2 dx

integral 1. space dy space equals space integral open parentheses sec squared straight x over 2 minus 1 close parentheses space dx comma space space or space space space straight y space equals space fraction numerator tan begin display style straight x over 2 end style over denominator begin display style 1 half end style end fraction minus straight x plus straight c

straight y space equals space 2 space tan straight x over 2 minus straight x plus straight c

which is the required solution.

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