How to do the ordinary differential equation?
 \[\dfrac{{dA}}{{dt}} = A = - \dfrac{\alpha }{{2m}}A \to A\left( t \right) = A\left( 0 \right){e^{ - \dfrac{\alpha }{{2m}}t}}\]

Hint: An ordinary differential equation (ODE) is an equation that involves some ordinary derivatives (as opposed to partial derivatives) of a function i.e., here we need to determine what function or functions satisfy the equation. In the given differential equation is a First-Order linear separable Ordinary Differential Equation, which we can rearrange terms and separate the variables to get the value of A.

Complete step-by-step answer:
Let us write the given data:
 \[\dfrac{{dA}}{{dt}} = A = - \dfrac{\alpha }{{2m}}A \to A\left( t \right) = A\left( 0 \right){e^{ - \dfrac{\alpha }{{2m}}t}}\] …………………… 1
Given that, the expression for A is:
 \[A = - \dfrac{\alpha }{{2m}}A\]
Then from equation 1 we have:
 \[ \Rightarrow \dfrac{{dA}}{{dt}} = - \dfrac{\alpha }{{2m}}A\]
This is a First-Order linear separable Ordinary Differential Equation, which we can rearrange terms and separate the variables to get:
 \[\int {\dfrac{1}{A}dA} = \int { - \dfrac{\alpha }{{2m}}} dt\]
Which is directly integrable, as we know that, \[\int {\dfrac{1}{x} = \log x} + C\] , hence we know that, \[\ln x = {\log _e}x\] , hence we get:
 \[ \Rightarrow \ln \left| A \right| = - \dfrac{\alpha }{{2m}}t + C\] ……………….. 2
If we use the initial condition as:
 \[A = A\left( 0 \right)\] when t=0
Then, we get equation 2 as:
 \[\ln \left| {A\left( 0 \right)} \right| = - \dfrac{\alpha }{{2m}}\left( 0 \right) + C\]
Simplifying the terms, we get:
 \[ \Rightarrow \ln \left| {A\left( 0 \right)} \right| = 0 + C\]
 \[ \Rightarrow \ln \left| {A\left( 0 \right)} \right| = C\] ……………………….. 3
Hence, leading to the Particular Solution i.e., substitute the value of C from equation 3 in equation 2 we get:
 \[\ln \left| A \right| = - \dfrac{\alpha }{{2m}}t + C\]
 \[ \Rightarrow \ln \left| A \right| = - \dfrac{\alpha }{{2m}}t + \ln \left| {A\left( 0 \right)} \right|\]
And, now assuming that A is a positive over its range, then:
 \[\ln A - \ln A\left( 0 \right) = - \dfrac{\alpha }{{2m}}t\]
Now, let's combine like terms together i.e., combining the terms of ln we get:
 \[ \Rightarrow \ln \left( {\dfrac{A}{{A\left( 0 \right)}}} \right) = - \dfrac{\alpha }{{2m}}t\] ……………………. 4
As, we know that the equation 4 is of the form \[\ln N = x\] , hence we get \[N = {e^x}\] , i.e., \[\ln N = x \Rightarrow N = {e^x}\] hence, applying this to the equation 4 we get:
 \[\ln \left( {\dfrac{A}{{A\left( 0 \right)}}} \right) = - \dfrac{\alpha }{{2m}}t\]
 \[ \Rightarrow \dfrac{A}{{A\left( 0 \right)}} = {e^{ - \dfrac{\alpha }{{2m}}t}}\] ……………………. 5
Therefore, shift the terms of equation 5, to get the value of A:
 \[ \Rightarrow A = A\left( 0 \right){e^{ - \dfrac{\alpha }{{2m}}t}}\]
So, the correct answer is “ \[ A = A\left( 0 \right){e^{ - \dfrac{\alpha }{{2m}}t}}\]”.

Note: To solve the ordinary differential equation, we must know; how to find the integration of the terms of the given function and then to identify the type of order i.e., first order, second order etc. and we must know all the basics involved to solve the sums related with ln function and exponential functions as ordinary differential equation involves some ordinary derivatives.