The order & degree of the following D.E. are m. n respectively.

∂³ϕ/∂x³ + (∂²ϕ/∂y²)(∂ϕ/∂x) + (∂²ϕ/∂x²)² + ∂²ϕ/∂y = 0  . The value of (m – n) is

The correct answer is 2 (Option 4, which corresponds to the value "2").

To find the value of $m-n$, we first need to determine the order ($m$) and degree ($n$) of the given partial differential equation (P.D.E.):

$$\frac{{\partial }^3\varphi }{\partial x^3}+\left(\frac{{\partial }^2\varphi }{\partial y^2}\right)\left(\frac{\partial \varphi }{\partial x}\right)+{\left(\frac{{\partial }^2\varphi }{\partial x^2}\right)}^2+\frac{\partial \varphi }{\partial y}=0$$

1. Finding the Order (m):
The order of a differential equation is defined as the order of the highest derivative present in the equation.
Looking at each term in the given equation:
- The first term $\frac{{\partial }^3\varphi }{\partial x^3}$ is a 3rd-order derivative.
- The term $\frac{{\partial }^2\varphi }{\partial y^2}$ is a 2nd-order derivative.
- The term $\frac{{\partial }^2\varphi }{\partial x^2}$ is a 2nd-order derivative.
- The terms $\frac{\partial \varphi }{\partial x}$ and $\frac{\partial \varphi }{\partial y}$ are 1st-order derivatives.
The highest derivative order present is 3. Therefore, the order of the differential equation, $m=3$.

2. Finding the Degree (n):
The degree of a differential equation is the power of the highest-order derivative when the equation is expressed as a polynomial in its derivatives (i.e., free from fractional powers or radicals in the derivatives).
Here, the highest-order derivative is $\frac{{\partial }^3\varphi }{\partial x^3}$.
The term containing this highest derivative is ${\left(\frac{{\partial }^3\varphi }{\partial x^3}\right)}^1$.
Since its power is 1, the degree of the differential equation, $n=1$.

3. Calculating (m - n):
Now we substitute the values of $m$ and $n$ to find $m-n$:
$$m-n=3-1=2$$

Thus, the value of $m-n$ is indeed 2.