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Vector Algebra

Question
CBSEENMA12033044

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

Solution
The given differential equation is
             (1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
or             left parenthesis 1 plus straight x right parenthesis space left parenthesis 1 plus straight y squared right parenthesis space dx space equals space minus left parenthesis 1 plus straight y right parenthesis thin space left parenthesis 1 plus straight x squared right parenthesis space dy
or           fraction numerator 1 plus straight y over denominator 1 plus straight y squared end fraction dy space equals space minus fraction numerator 1 plus straight x over denominator 1 plus straight x squared end fraction dx space space space rightwards double arrow space space space integral fraction numerator 1 plus straight y over denominator 1 plus straight y squared end fraction dy space equals space minus integral fraction numerator 1 plus straight x over denominator 1 plus straight x squared end fraction dx
therefore space space space space integral fraction numerator 1 over denominator 1 plus straight y squared end fraction dy plus 1 half integral fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy space equals space minus integral fraction numerator 1 over denominator 1 plus straight x squared end fraction dx minus 1 half integral fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction dx
therefore space space space space space tan to the power of negative 1 end exponent straight y space plus space 1 half space log space left parenthesis 1 plus straight y squared right parenthesis space equals space minus tan to the power of negative 1 end exponent straight x space minus space 1 half log left parenthesis 1 plus straight x squared right parenthesis space plus straight c
or         tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y plus 1 half open square brackets log space left parenthesis 1 plus straight x squared right parenthesis space plus space log space left parenthesis 1 plus straight y squared right parenthesis close square brackets space equals space straight c
or            tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight y plus 1 half log space open square brackets left parenthesis 1 plus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis close square brackets space equals space straight c space is space the space required space solution. space