The "order" of the following ordinary differential equation is $\qquad$ .

$$ \frac{d^3 y}{d x^3}+\left(\frac{d^2 y}{d x^2}\right)^6+\left(\frac{d y}{d x}\right)^4+y=0 $$

A partial differential equation

$$\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0$$

is defined for the two-dimensional field $T: T(x, y)$, inside a planar square domain of size 2 m × 2 m. Three boundary edges of the square domain are maintained at value $T = 50$, whereas the fourth boundary edge is maintained at $T = 100$.

The value of $T$ at the center of the domain is

Consider two Ordinary Differential Equations (ODEs):

P: $ \dfrac{dy}{dx} = \dfrac{x^4 + 3x^2 y^2 + 2y^4}{x^3 y} $

Q: $ \dfrac{dy}{dx} = -\dfrac{y^2}{x^2} $

Which one of the following options is CORRECT?