The sum of the order and degree of the differential equation $\frac{d^4 y}{d x^4}=\{c+(\frac{d y}{d x})^2\}^{\frac{3}{2}}$ is

Consider the following differential equations.
$D_1: y=4 \frac{dy}{dx}+3x \frac{dx}{dy}$
$D_2: \frac{d^2y}{dx^2}=\left(3+\left(\frac{dy}{dx}\right)^2\right)^{\frac{4}{3}}$
$D_3: \left[1+\left(\frac{dy}{dx}\right)\right]^2=\left(\frac{dy}{dx}\right)^2$
The ratio of the sum of the orders of $D_1, D_2$ and $D_3$ to the sum of their degrees is

For the differential equation $\left[1-\left(\frac{dy}{dx}\right)^2\right]^{5/2} = 8 \frac{d^2y}{dx^2}$,the order and degree are:

The order and degree of the differential equation $\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2}=0$ are respectively:

If $m$ and $n$ are the order and degree of the differential equation $\left(\frac{d^{2} y}{d x^{2}}\right)^{5}+4 \cdot \frac{\left(\frac{d^{2} y}{d x^{2}}\right)^{3}}{\left(\frac{d^{3} y}{d x^{3}}\right)}+\left(\frac{d^{3} y}{d x^{3}}\right)=x^{2}-1$,then:

Determine the order and degree (if defined) of the differential equation $\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$.