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The question is in the form of standard linear differential equation i.e. $\frac{dy}{dx}+yP\left ( x \right )=Q\left ( x \right )$,

for which the integrating factor is   $I=e^{\int P\left ( x \right )dx}$.

For the given question, converting to standard form,  $\frac{dy}{dx}+y\left ( -\frac{1}{x} \right )=1$

So, the Integrating factor would be:   $e^{\int \left ( -\frac{1}{x} \right )dx}=\frac{1}{x}$.

$\therefore$  Solution to such a Differential equation is :   $ye^{\int P\left ( x \right )dx}=\int \left \{ Q\left ( x \right )e^{\int P\left ( x \right )dx} \right \}dx$

Now, using the above solution definition, $y\left ( \frac{1}{x} \right )=\int \left ( 1 \right )\left ( \frac{1}{x} \right )dx$

$\Rightarrow$     $\frac{y}{x}=log_{e}x+c$     $\Rightarrow$  $y=x\left ( log_{e}x+c \right )$

 

Option A is the correct answer.

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