The differential equation representing the family of curves $y = Ax + \frac{B}{x}$, $x \neq 0$ where A and B are arbitrary constants, is given by

The family of curves is $y = Ax + \frac{B}{x}$ where A and B are arbitrary constants.

Since there are 2 arbitrary constants, differentiate twice to eliminate them.

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Given: $y = Ax + \frac{B}{x}$

Differentiating with respect to x:

$\frac{dy}{dx} = A - \frac{B}{x^2}$

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Differentiating again:

$\frac{d^2y}{dx^2} = \frac{2B}{x^3}$

From this:

$B = \frac{x^3}{2} \cdot \frac{d^2y}{dx^2}$

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From the first derivative:

$A = \frac{dy}{dx} + \frac{B}{x^2}$

Substituting the value of B:

$A = \frac{dy}{dx} + \frac{1}{x^2} \cdot \frac{x^3}{2}\frac{d^2y}{dx^2}$

$A = \frac{dy}{dx} + \frac{x}{2}\frac{d^2y}{dx^2}$

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Substituting A and B in the original equation $y = Ax + \frac{B}{x}$:

$y = \left(\frac{dy}{dx} + \frac{x}{2}\frac{d^2y}{dx^2}\right) \cdot x + \frac{1}{x} \cdot \frac{x^3}{2}\frac{d^2y}{dx^2}$

$y = x\frac{dy}{dx} + \frac{x^2}{2}\frac{d^2y}{dx^2} + \frac{x^2}{2}\frac{d^2y}{dx^2}$

$y = x\frac{dy}{dx} + x^2\frac{d^2y}{dx^2}$

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Therefore, the differential equation is:

$x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = 0$