if $y_1 = \ln x$ and $y_2 = x + \ln x$ are particular solution, then the difference $y_2 - y_1 = x$ is a solution to the associated homogeneous equation. now we have two linearly independent solutions $y = x, y = e^{2x}.$ they satisfy
$$\frac{d^2y}{dx^2} = 0 \text{ and } \frac{dy}{dx} - 2y = 0$$ now we can form a third order differential equation
$$0 = \frac{d^2}{dx^2}(\frac{d}{dx} - 2)y = \frac{d^3y}{dx^3} - 2\frac{d^2y}{dx^2}$$
now we will get the forced equation and the force so that $\ln x$ is a particular solution.
$(\frac{d^3}{dx^3} -2\frac{d^2}{dx^2}) \ln x = {2 \over x^3} + {2 \over x^2} =
{2(x+1) \over x^3}.$ here is the equation satisfying all the requirements:
$$\frac{d^3y}{dx^3} - 2\frac{d^2y}{dx^2} = {2(x+1) \over x^3}.$$