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    Maths-
    General
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    Question

    The solution of fraction numerator d to the power of 2 end exponent y over denominator d x to the power of 2 end exponent end fraction equals sec to the power of 2 end exponent invisible function application x plus x e to the power of x end exponent is

    1. y equals log invisible function application left parenthesis sec invisible function application x right parenthesis plus left parenthesis x minus 2 right parenthesis e to the power of x end exponent plus c subscript 1 end subscript x plus c subscript 2 end subscript    
    2. y equals log invisible function application left parenthesis sec invisible function application x right parenthesis plus left parenthesis x plus 2 right parenthesis e to the power of x end exponent plus c subscript 1 end subscript x plus c subscript 2 end subscript    
    3. y equals log invisible function application left parenthesis sec invisible function application x right parenthesis minus left parenthesis x plus 2 right parenthesis e to the power of x end exponent plus c subscript 1 end subscript x plus c subscript 2 end subscript    
    4. None of these    

    The correct answer is: y equals log invisible function application left parenthesis sec invisible function application x right parenthesis plus left parenthesis x minus 2 right parenthesis e to the power of x end exponent plus c subscript 1 end subscript x plus c subscript 2 end subscript

      fraction numerator d to the power of 2 end exponent y over denominator d x to the power of 2 end exponent end fraction equals sec to the power of 2 end exponent invisible function application x plus x e to the power of x end exponent
      On integrating, fraction numerator d y over denominator d x end fraction equals tan invisible function application x plus x e to the power of x end exponent minus e to the power of x end exponent plus c subscript 1 end subscript
      Again, y equals log invisible function application left parenthesis sec invisible function application x right parenthesis plus x e to the power of x end exponent minus e to the power of x end exponent minus e to the power of x end exponent plus c subscript 1 end subscript x plus c subscript 2 end subscript
      Thus required solution is
      y equals log invisible function application left parenthesis sec invisible function application x right parenthesis plus left parenthesis x minus 2 right parenthesis e to the power of x end exponent plus c subscript 1 end subscript x plus c subscript 2 end subscript.

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