The solution of the differential equation $x^2 \dfrac{dy}{dx} - xy = 1$ is...

The given differential equation is $x^2 \dfrac{dy}{dx} - xy = 1$

Dividing by $x^2$, we get:

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

This is a linear differential equation of the form $\dfrac{dy}{dx} + Py = Q$, where $P = -\dfrac{1}{x}$ and $Q = \dfrac{1}{x^2}$

Integrating Factor (IF) = $e^{\int P dx} = e^{\int -\dfrac{1}{x} dx} = e^{-\ln x} = \dfrac{1}{x}$

The solution is given by:

$y \cdot (\text{IF}) = \int Q \cdot (\text{IF}) dx + c$

$y \left(\dfrac{1}{x}\right) = \int \dfrac{1}{x^2} \left(\dfrac{1}{x}\right) dx + c$

$\dfrac{y}{x} = \int x^{-3} dx + c$

$\dfrac{y}{x} = \dfrac{x^{-2}}{-2} + c$

$\dfrac{y}{x} = -\dfrac{1}{2x^2} + c$

Multiplying the entire equation by $2x^2$, we get:

$2xy = -1 + 2cx^2$

$2xy - 2cx^2 + 1 = 0$

Answer: $2xy - 2cx^2 + 1 = 0$