What is the degree of the following differential equation?
$$x=\sqrt{1+\frac{d^{2}y}{d{x}^2}}$$

To find the degree, we must first express the differential equation in a form free from radicals regarding the derivatives.Squaring both sides of the given equation:
$$x^2=1+\frac{d^{2}y}{d{x}^2}$$
Rearranging the terms:
$${\left(\frac{d^{2}y}{d{x}^2}\right)}^1=x^2-1$$
The highest order derivative present is $\frac{d^{2}y}{d{x}^2}$, so the order is $2$.The power of the highest order derivative is $1$. Therefore, the degree of the differential equation is $1$.