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Question

The solution of the differential equation $fraction numerator d y over denominator d x end fraction plus fraction numerator y over denominator x end fraction equals x to the power of 2 end exponent$ is

$4 x y = x^{4} + c$
$x y = x^{4} + c$
$\frac{1}{4} x y = x^{4} + c$
$x y = 4 x^{4} + c$

The correct answer is: $4 x y equals x to the power of 4 end exponent plus c$

The given equation $fraction numerator d y over denominator d x end fraction plus fraction numerator y over denominator x end fraction equals x to the power of 2 end exponent$ is of the form
$fraction numerator d y over denominator d x end fraction plus P y equals Q$ . So, I.F.= $e to the power of not stretchy integral fraction numerator 1 over denominator x end fraction d x end exponent equals e to the power of log invisible function application x end exponent equals x$
Hence required solution $x y equals not stretchy integral x. x to the power of 2 end exponent d x plus c$
Þ $x y equals fraction numerator x to the power of 4 end exponent over denominator 4 end fraction plus c$ Þ $4 x y equals x to the power of 4 end exponent plus c$ .

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