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Vector Algebra

Question
CBSEENMA12033073

Solve the differential equation;
x (1 + y2 ) dx – y (1 + x2 ) dy = 0  given that y = 0 when x = 1. 

Solution
The given differential equation is
           x (1 + y2 ) dx – y (1 + x2 ) dy = 0 
or              y (1 + x2 ) dy = 0 =    x (1 + y2 ) dx
therefore space space space space space space space space space space space space fraction numerator straight y over denominator 1 plus straight y squared end fraction dy space equals space fraction numerator straight x over denominator 1 plus straight x squared end fraction dx
rightwards double arrow space space space space space space space space space space fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy space equals space fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction dx
Integrating, integral fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy space equals space integral fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction dx
therefore space space space space space space space space space space space space space log space left parenthesis 1 plus straight y squared right parenthesis space equals space log space left parenthesis 1 plus straight x squared right parenthesis space plus space log space straight c
therefore space space space space space space space space log space left parenthesis 1 plus straight y squared right parenthesis space equals space log space left square bracket straight c space left parenthesis 1 plus straight x squared right parenthesis right square bracket
rightwards double arrow space space space space space space space space space space space space space space space space space space space space space space space space 1 plus straight y squared space equals space straight c space left parenthesis 1 plus straight x squared right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Now   y = 0 when x = 1
therefore space space space space space space space space space 1 plus 0 space equals space straight c space left parenthesis 1 plus 1 right parenthesis space space or space space space space 1 space equals space 2 straight c space space space space rightwards double arrow space space straight c space equals space 1 half
therefore space space space space from space left parenthesis 1 right parenthesis comma space space space space space space space 1 plus straight y squared space equals space 1 half left parenthesis 1 plus straight x squared right parenthesis
which is required solution.