Given family of curves

y^2 = 2cigl(x+\sqrt c\bigr), \qquad c>0

Here $c$ is the single arbitrary parameter, so the differential equation should be of order 1 after eliminating $c$.

Differentiate with respect to $x$

Since $c$ is a parameter (constant), differentiating gives

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

Hence,

$c = y\frac{dy}{dx}$

Let $y' = \dfrac{dy}{dx}$. Then

$c = yy'$

Substitute back into the original family

Original equation:

$y^2 = 2c\left(x+\sqrt c\right)$

Replacing $c$ by $yy'$:

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

Remove the radical to express as a polynomial in $y'$

Divide both sides by $2yy'$ (assuming nontrivial branch where this is valid; order/degree are determined from the resulting polynomial form):

$\frac{y}{2y'} = x+\sqrt{yy'}$

So,

$\sqrt{yy'} = \frac{y}{2y'} - x$

Squaring:

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

Multiply by $4(y')^2$:

$4yy'^3 = (y-2xy')^2$

Expanding the right side:

$4yy'^3 = y^2 - 4xyy' + 4x^2 y'^2$

Rearranging:

$4yy'^3 - 4x^2 y'^2 + 4xyy' - y^2 = 0$

Determine order and degree

• The highest derivative present is $y'$, so the order is $1$.
• The equation is polynomial in $y'$, and the highest power of $y'$ is $3$, so the degree is $3$.

Conclusion

The differential equation is of:

$\boxed{\text{order }1,\ \text{degree }3}$

Therefore, the correct option is:

$\boxed{\text{C}}$