Sum of the squares of the order and degree (if defined) of a differential equation \(2 y^{\prime}+\left(y^{\prime \prime}\right)^2=\sqrt{y^{\prime \prime}-3}\) is

Given differential equation is \(2 y^{\prime} + (y^{\prime \prime})^2 = \sqrt{y^{\prime \prime} - 3}\)

To find the degree, we must express the differential equation as a polynomial in its derivatives. Squaring both sides, we get:

\((2 y^{\prime} + (y^{\prime \prime})^2)^2 = y^{\prime \prime} - 3\)

\(4 (y^{\prime})^2 + 4 y^{\prime} (y^{\prime \prime})^2 + (y^{\prime \prime})^4 = y^{\prime \prime} - 3\)

\((y^{\prime \prime})^4 + 4 y^{\prime} (y^{\prime \prime})^2 - y^{\prime \prime} + 4 (y^{\prime})^2 + 3 = 0\)

The highest order derivative present in the equation is \(y^{\prime \prime}\), so the order is \(2\).

The highest power of the highest order derivative \(y^{\prime \prime}\) is \(4\), so the degree is \(4\).

Sum of the squares of the order and degree is \(2^2 + 4^2 = 4 + 16 = 20\).

Answer: \(20\)