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

3

The correct answer is Option (3) → 3

Given differential equation:

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

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

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

Left-hand side:

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

Right-hand side:

$ 4 + 4 \left( \frac{dy}{dx} \right)^2 $

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

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

Now this is a polynomial equation in $ \frac{dy}{dx} $.

Order: 1 (highest order derivative is first order)

Degree: 2 (highest power of the first derivative after removing radicals is 2)

Sum of order and degree: $ 1 + 2 = 3 $