The sum of order and degree of the differential equation $y = x\frac{dy}{dx} + 2\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$ is

The given differential equation is:

$y = x\frac{dy}{dx} + 2\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$

The order of a differential equation is the highest derivative present. The highest derivative here is $\frac{dy}{dx}$ (first derivative).

Order $= 1$

The degree is the power of the highest derivative after the equation is expressed as a polynomial in derivatives. The equation currently has a square root containing $\frac{dy}{dx}$, so it must be converted to polynomial form.

$y - x\frac{dy}{dx} = 2\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$

$\left(y - x\frac{dy}{dx}\right)^2 = 4\left(1 + \left(\frac{dy}{dx}\right)^2\right)$

$y^2 - 2xy\frac{dy}{dx} + x^2\left(\frac{dy}{dx}\right)^2 = 4 + 4\left(\frac{dy}{dx}\right)^2$

$y^2 - 2xy\frac{dy}{dx} + x^2\left(\frac{dy}{dx}\right)^2 - 4\left(\frac{dy}{dx}\right)^2 - 4 = 0$

$y^2 - 2xy\frac{dy}{dx} + (x^2 - 4)\left(\frac{dy}{dx}\right)^2 - 4 = 0$

The equation is now polynomial in $\frac{dy}{dx}$. The highest power of $\frac{dy}{dx}$ is $2$.

Degree $= 2$

Sum $=$ Order $+$ Degree

Sum $= 1 + 2$

Sum $= 3$

Therefore, the sum of order and degree is $3$.