The solution of the differential equation $\frac{dy}{dx}-\frac{y}{x}=2 \log_ex$

$y=x((\log_ex)^2 + C)$: C is an arbitrary constant

The correct answer is Option (3) → $y=x((\log_ex)^2 + C)$: C is an arbitrary constant

Given differential equation: $\frac{dy}{dx}-\frac{y}{x}=2\log_e x$

It is a linear equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$ with $P(x)=-\frac{1}{x}$ and $Q(x)=2\log_e x$.

Integrating factor (I.F.) $= e^{\int P(x)dx}=e^{-\int \frac{1}{x}dx}=e^{-\log x}=\frac{1}{x}$

Multiplying both sides by $\frac{1}{x}$:

$\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^{2}}=\frac{2\log x}{x}$

$\Rightarrow \frac{d}{dx}\!\left(\frac{y}{x}\right)=\frac{2\log x}{x}$

Integrate both sides:

$\frac{y}{x}=\int \frac{2\log x}{x}dx + C$

Let $t=\log x \Rightarrow dt=\frac{dx}{x}$

$\frac{y}{x}=2\int t\,dt + C = 2\cdot\frac{t^{2}}{2}+C=(\log x)^{2}+C$

$\Rightarrow y=x((\log_e x)^{2}+C)$

$y=x((\log_e x)^{2}+C)$