The difference between degree and order of a differential equation that represents the family of curves given by $$y^2 = a\left(x + \frac{\sqrt{a}}{2}\right), a > 0$$ is ______.

The family of curves is given by $$y^2 = a\left(x + \frac{\sqrt{a}}{2}\right)$$, where $$a > 0$$ is the parameter to be eliminated.

Expanding: $$y^2 = ax + \frac{a^{3/2}}{2}$$.

Differentiating both sides with respect to $$x$$: $$2yy' = a$$, so $$a = 2yy'$$.

Substituting back into the original equation: $$y^2 = 2yy' \cdot x + \frac{(2yy')^{3/2}}{2}$$.

$$y^2 = 2xyy' + \frac{2^{3/2}(yy')^{3/2}}{2} = 2xyy' + \sqrt{2}(yy')^{3/2}$$

Dividing both sides by $$y$$ (assuming $$y \neq 0$$): $$y = 2xy' + \sqrt{2}y^{1/2}(y')^{3/2}$$.

To clear the fractional powers, we isolate the radical term: $$y - 2xy' = \sqrt{2}y^{1/2}(y')^{3/2}$$.

Squaring both sides: $$(y - 2xy')^2 = 2y(y')^3$$.

This is the differential equation representing the family. The order is 1 (highest derivative is $$y'$$) since we eliminated one parameter $$a$$. The degree is the highest power of the highest order derivative when the equation is a polynomial in derivatives. Expanding the left side gives terms involving $$(y')^2$$ and lower, while the right side has $$(y')^3$$. So the degree is 3.

The difference between degree and order is $$3 - 1 = 2$$.