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

The differential equation is $\frac{dy}{dx} - \frac{y}{x} = 2\log_e x$

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

---

The integrating factor is given by $IF = e^{\int P(x)dx}$

$IF = e^{\int -\frac{1}{x}dx}$

$IF = e^{-\log_e x}$

$IF = e^{\log_e x^{-1}}$

$IF = x^{-1}$

$IF = \frac{1}{x}$

---

Multiplying the entire equation by $\frac{1}{x}$:

$\frac{1}{x} \cdot \frac{dy}{dx} - \frac{1}{x} \cdot \frac{y}{x} = \frac{1}{x} \cdot 2\log_e x$

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

---

The left side is the derivative of $\frac{y}{x}$ by the quotient rule:

$\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{x\frac{dy}{dx} - y \cdot 1}{x^2} = \frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2}$

The equation becomes:

$\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{2\log_e x}{x}$

---

Integrating both sides:

$\frac{y}{x} = \int \frac{2\log_e x}{x} dx$

Let $u = \log_e x$, then $\frac{du}{dx} = \frac{1}{x}$:

$\int \frac{2\log_e x}{x} dx = \int 2u \, du$

$= 2 \cdot \frac{u^2}{2} + C$

$= u^2 + C$

$= (\log_e x)^2 + C$

---

Therefore:

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

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

where $C$ is an arbitrary constant.