Determine the order and degree (if defined) of the differential equation $y^{\prime} + 5y = 0$.

Determine the order and degree (if defined) of the differential equation: $(\frac{d^3y}{dx^3})^2 + (\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^4 + y^5 = 0$.

If $y=a^3 e^{b^2 x+c}$ is the general solution of a differential equation,where $a$ and $c$ are arbitrary constants and $b$ is a fixed constant,then the order of the differential equation is

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:

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

Find the order and degree,if defined,of the following differential equation:
$x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0$