1. Given family of curves

   $y^2=a\left(x+\frac{\sqrt a}{2}\right), \qquad a>0$

   We need the differential equation representing this family, and then find:

   $\text{degree} - \text{order}$

2. Differentiate the family with respect to $x$

   Since $a$ is a parameter (constant for a given curve), differentiating gives

   $2y\frac{dy}{dx}=a$

   So,

   $a=2y\,y'$

   where $y' = \dfrac{dy}{dx}$.

3. Eliminate the parameter $a$

   The original equation contains both $a$ and $\sqrt a$. Since

   $a=2yy',$

   we also have

   $\sqrt a = \sqrt{2yy'}$

   Substitute into the original family:

   $y^2 = a\left(x+\frac{\sqrt a}{2}\right)$

   $y^2 = (2yy')\left(x+\frac{\sqrt{2yy'}}{2}\right)$

   Simplify:

   $y^2 = 2xyy' + yy'\sqrt{2yy'}$

   Divide by $y$ (for the differential equation form):

   $y = 2xy' + y'\sqrt{2yy'}$

4. Remove the radical to obtain polynomial form in derivatives

   $y - 2xy' = y'\sqrt{2yy'}$

   Squaring both sides:

   $\left(y-2xy'\right)^2 = (y')^2(2yy')$

   $\left(y-2xy'\right)^2 = 2y(y')^3$

   This is the required differential equation.

5. Find order and degree

   The highest order derivative present is $y'$, so

   $\text{order} = 1$

   The equation is polynomial in $y'$:

   $\left(y-2xy'\right)^2 = 2y(y')^3$

   Highest power of $y'$ is $3$, so

   $\text{degree} = 3$

6. Difference between degree and order

   $\text{degree} - \text{order} = 3-1=2$

Final Answer

$\boxed{2}$