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Differential Equations

Question

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Solve:
$cos space left parenthesis straight x plus straight y right parenthesis space dy over dx space equals space 1.$

Solution

The given differential equation is

dy over dx space equals space fraction numerator 1 over denominator cos space left parenthesis straight x plus straight y right parenthesis end fraction

...(1)
Put x+y = t,

therefore space space space 1 plus dy over dx space equals space dt over dx space space or space space dy over dx space equals space dt over dx minus 1

therefore space space from space left parenthesis 1 right parenthesis comma space space space dt over dx minus 1 space equals space fraction numerator 1 over denominator cos space straight t end fraction space space space space space space space space space rightwards double arrow space space space space space dt over dx space equals space 1 plus fraction numerator 1 over denominator cos space straight t end fraction therefore space space space space dt over dx equals space fraction numerator 1 plus cos space straight t over denominator cos space straight t end fraction space space space rightwards double arrow space space space space space fraction numerator cos space straight t over denominator 1 plus space cos space straight t end fraction dt space equals space dx therefore space space space integral fraction numerator cos space straight t over denominator 1 plus space cos space straight t end fraction dt space equals space integral 1. space dx space space space rightwards double arrow space space space integral fraction numerator left parenthesis 1 plus cost right parenthesis minus 1 over denominator 1 plus cost end fraction dt space equals space integral 1. space dx therefore space space space space integral open parentheses 1 minus fraction numerator 1 over denominator 1 plus cost end fraction close parentheses dt space equals space integral 1. space dx space space space rightwards double arrow space space space integral open parentheses 1 minus fraction numerator 1 over denominator 2 space cos squared begin display style straight t over 2 end style end fraction close parentheses dt space equals space integral 1. space dx therefore space space space integral open parentheses 1 minus 1 half sec squared straight t over 2 close parentheses space dt space equals space integral 1. space dx space space space space rightwards double arrow space space space straight t minus tan straight t over 2 space equals space straight x plus straight c apostrophe therefore space space space space straight x plus straight y minus tan fraction numerator straight x plus straight y over denominator 2 end fraction space equals space straight x plus straight c apostrophe space space space space rightwards double arrow space space space tan fraction numerator straight x plus straight y over denominator 2 end fraction space equals space straight y plus straight c comma space which space is space required space solution. comma space

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