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Vector Algebra

Question
CBSEENMA12033043

Solve:
x (1 + y2) dx + y (1 + x2)dy = 0  

Solution

The given differential equation is
x(1 + y2) dx + y(1 + x2)dy = 0  
therefore space space space space straight y left parenthesis 1 plus straight x squared right parenthesis space dy equals space minus straight x left parenthesis 1 plus straight y squared right parenthesis space dx
therefore space space space space space fraction numerator straight y over denominator 1 plus straight x squared end fraction dy space equals space minus fraction numerator straight x over denominator 1 plus straight x squared end fraction dx
Integrating, integral fraction numerator straight y over denominator 1 plus straight y squared end fraction dy space equals space minus integral fraction numerator straight x over denominator 1 plus straight x squared end fraction dx
therefore space space space space integral fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy plus integral fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction dx space equals space 0
therefore space space space log space left parenthesis 1 plus straight y squared right parenthesis plus space log space left parenthesis 1 plus straight x squared right parenthesis space equals space log space straight c
therefore space space log space left parenthesis 1 plus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis space equals space log space straight c
therefore space space left parenthesis 1 plus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis space equals straight c
which is required solution.