Which of the following is a second order differential equation?

The correct option is y’y” + y = sin x.

To understand why this is the correct answer, let us first define the order of a differential equation.
The order of a differential equation is the order of the highest derivative present in the equation.

Let us analyze each of the given options by identifying their highest order derivatives:

1. In the equation ${\left(y\prime \right)}^2+x=y^2$, the highest derivative is the first derivative, $y\prime$ (or $\frac{dy}{dx}$). The exponent of $2$ represents the power (degree) of the derivative, not its order. Thus, this is a first-order differential equation.

2. In the equation $y\prime y″+y=sinx$, the highest derivative is the second derivative, $y″$ (or $\frac{d^{2}y}{d{x}^2}$). Since the highest derivative present is of order two, this is a second-order differential equation.

3. In the equation $y‴+{\left(y″\right)}^2+y=0$, the highest derivative is the third derivative, $y‴$ (or $\frac{d^{3}y}{d{x}^3}$). Therefore, this is a third-order differential equation.

4. In the equation $y\prime =y^2$, the highest derivative is the first derivative, $y\prime$. Thus, this is a first-order differential equation.

Consequently, the only second-order differential equation among the choices is indeed $y\prime y″+y=sinx$.