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Vector Algebra

Question
CBSEENMA12033055

Solve:
 left parenthesis 1 plus straight e to the power of 2 straight x end exponent right parenthesis space dy space plus space left parenthesis 1 plus straight y squared right parenthesis space straight e to the power of straight x space dx space equals space 0

Solution
The given differential equation is
      left parenthesis 1 plus straight e to the power of 2 straight x end exponent right parenthesis space dy space plus space left parenthesis 1 plus straight y squared right parenthesis space straight e to the power of straight x space dx space equals space 0 space space or space space left parenthesis 1 plus straight e to the power of 2 straight x end exponent right parenthesis space dy space equals space minus left parenthesis 1 plus straight y squared right parenthesis space straight e to the power of straight x space dx
Separating the variables, we get,
                 fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space minus fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx
Integrating, integral fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space minus integral fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx             ...(1)
Let          straight I space equals integral fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx
Put straight e to the power of straight x space equals space straight t comma space space space space therefore space space space straight e to the power of straight x dx space equals space dt
therefore space space space space space space space space space straight I space equals space integral fraction numerator 1 over denominator 1 plus straight t squared end fraction dt space space equals space tan to the power of negative 1 end exponent straight t plus straight c subscript 1 space equals space tan to the power of negative 1 end exponent left parenthesis straight e to the power of straight x right parenthesis space plus space straight c subscript 1
therefore space space space space from space left parenthesis 1 right parenthesis comma space space space tan to the power of negative 1 end exponent straight y space equals space minus tan to the power of negative 1 end exponent left parenthesis straight e to the power of straight x right parenthesis space plus space straight c subscript 1
therefore space space space space tan to the power of negative 1 end exponent straight y space plus space tan to the power of negative 1 end exponent left parenthesis straight e to the power of straight x right parenthesis space equals space straight c subscript 1
therefore space space space tan to the power of negative 1 end exponent open parentheses fraction numerator straight y plus straight e to the power of straight x over denominator 1 minus ye to the power of straight x end fraction close parentheses space equals space straight c subscript 1
therefore space space space fraction numerator straight y plus straight e to the power of straight x over denominator 1 minus ye to the power of straight x end fraction space equals space straight c comma space space space where space straight c space equals space tan space straight c subscript 1
This is the required solution.