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Maths-

General

Easy

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

$y = \log (\sec x) + (x - 2) e^{x} + c_{1} x + c_{2}$
$y = \log (\sec x) + (x + 2) e^{x} + c_{1} x + c_{2}$
$y = \log (\sec x) - (x + 2) e^{x} + c_{1} x + c_{2}$
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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