Degree of differential equation $\sqrt{1+\frac{d^2y}{dx^2}}=x+\frac{dy}{dx}$

Question

Degree of the differential equation $\sqrt{1+\frac{d^2y}{dx^2}}=x+\frac{dy}{dx}$

Degree of the differential equation $\sqrt{1+\frac{d^2y}{dx^2}}=x+\frac{dy}{dx}$ is said to be "not defined" in my reference.

Doubt 1

What if I do: $$ \color{blue}{\sqrt{1+\frac{d^2y}{dx^2}}=x+\frac{dy}{dx}}\implies 1+\frac{d^2y}{dx^2}=\bigg[x+\frac{dy}{dx}\bigg]^2 $$ Now can I say the degree is $1$. If not, is it because squaring both sides add more information into the equation that both the differential equations are not exactly the same ?

Doubt 2

Thanx @farruhota for the link: How to find degree of a differential equation or check page 4 of ordinary differential equations by E. L. Ince

If I'm allowed to square both sides and say that the degree is $1$, as in the link, can I also square again and say the degree is $2$ right ?

Answer 1

The order is the greatest $n$ for which an $n$th derivative is present, while the degree is the highest power of such maximal-order derivatives, after the equation is written in rational form with all exponents integers. The equation $1+y''=x^2+2xy'+y'^2$ has order $2$, degree $1$.