The differential equation found by the elimination of the arbitrary constant \[{\rm{K}}\] from the equation\[y = (x + K){e^{ - x}}\]is
A. \[\frac{{dy}}{{dx}} - y = {e^{ - x}}\]
B. \[\frac{{dy}}{{dx}} - y{e^x} = 1\]
C. \[\frac{{dy}}{{dx}} + y{e^x} = 1\]
D. \[\frac{{dy}}{{dx}} + y = {e^{ - x}}\]

Hint:
Before attempting to answer the issue, it is important to understand what a differential equation is and what the arbitrary constants are in all equations. Following that, we will compare y to x and discriminate between them. From this, we shall have an equation in the form of x, y, and y's derivative. We shall once more distinguish the resulting equation with regard to x. The second derivative of \[y,x\], and \[y\] will be given by this equation as well. We will then calculate the relationship between\[y\]and\[y\]'s second derivative from this.
Formula used:
\[y = {x^n}\]
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = n{x^{n-1}}\]
\[y = {e^{-x}}\]
\[\frac{{dy}}{{dx}} ={-e^{-x}}\]

Complete step-by-step solution:
We have been given the equation of the arbitrary constant \[{\rm{K}}\]
 \[y = (x +K){e^{ - x}}............(1)\]
On differentiating the equation (1) with respect to \[x\], we obtain
\[\frac{{dy}}{{dx}} = {e^{ - x}}[1] + (x +K){e^{ - x}}( - 1)\]
Simplify the above equation to make it less complicated:
\[ \Rightarrow \frac{{dy}}{{dx}} = {e^{ - x}} - y...........(2)\]
Let’s rearrange the equation (2) explicitly to have \[{e^{ - x}}\]on one side, we get
\[ \Rightarrow \frac{{dy}}{{dx}} + y = {e^{ - x}}\]
Therefore, the differential equation found by the elimination of the arbitrary constant \[{\rm{K}}\] from the equation \[y = (x + K){e^{ - x}}\] is \[\frac{{dy}}{{dx}} + y = {e^{ - x}}\]
Hence, the option D is correct.
Note:
Eliminating the arbitrary constants from the equation before forming a differential equation is a need. In general, we differentiate the given equation so many times that the number of arbitrary constants is equal to the number of times. The idea is that we need to remove the arbitrary constant from the equation, not that we need to discover the differential equation. The differentiation may vary the same amount of times.