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Question

The solution of the differential equation x left parenthesis e to the power of 2 y end exponent minus 1 right parenthesis d y plus left parenthesis x to the power of 2 end exponent minus 1 right parenthesis e to the power of y end exponent d x equals 0 is

  1. e to the power of y end exponent plus e to the power of negative y end exponent equals log invisible function application x minus fraction numerator x to the power of 2 end exponent over denominator 2 end fraction plus c e    
  2. e to the power of y end exponent minus e to the power of negative y end exponent equals log invisible function application x minus fraction numerator x to the power of 2 end exponent over denominator 2 end fraction plus c    
  3. e to the power of y end exponent plus e to the power of negative y end exponent equals log invisible function application x plus fraction numerator x to the power of 2 end exponent over denominator 2 end fraction plus c    
  4. None of these    

The correct answer is: e to the power of y end exponent plus e to the power of negative y end exponent equals log invisible function application x minus fraction numerator x to the power of 2 end exponent over denominator 2 end fraction plus c e


    x left parenthesis e to the power of 2 y end exponent minus 1 right parenthesis d y plus left parenthesis x to the power of 2 end exponent minus 1 right parenthesis e to the power of y end exponent d x equals 0
    Þ not stretchy integral fraction numerator e to the power of 2 y end exponent minus 1 over denominator e to the power of y end exponent end fraction d y equals not stretchy integral fraction numerator 1 minus x to the power of 2 end exponent over denominator x end fraction d xÞ e to the power of y end exponent plus e to the power of negative y end exponent equals log invisible function application x minus fraction numerator x to the power of 2 end exponent over denominator 2 end fraction plus c.

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