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Question: How to do the ordinary differential equation? \[\dfrac{{dA}}{{dt}} = A = - \dfrac{\alpha }{{2m}}...

How to do the ordinary differential equation?
dAdt=A=α2mAA(t)=A(0)eα2mt\dfrac{{dA}}{{dt}} = A = - \dfrac{\alpha }{{2m}}A \to A\left( t \right) = A\left( 0 \right){e^{ - \dfrac{\alpha }{{2m}}t}}

Explanation

Solution

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:
dAdt=A=α2mAA(t)=A(0)eα2mt\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=α2mAA = - \dfrac{\alpha }{{2m}}A
Then from equation 1 we have:
dAdt=α2mA\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:
1AdA=α2mdt\int {\dfrac{1}{A}dA} = \int { - \dfrac{\alpha }{{2m}}} dt
Which is directly integrable, as we know that, 1x=logx+C\int {\dfrac{1}{x} = \log x} + C , hence we know that, lnx=logex\ln x = {\log _e}x , hence we get:
lnA=α2mt+C\Rightarrow \ln \left| A \right| = - \dfrac{\alpha }{{2m}}t + C ……………….. 2
If we use the initial condition as:
A=A(0)A = A\left( 0 \right) when t=0
Then, we get equation 2 as:
lnA(0)=α2m(0)+C\ln \left| {A\left( 0 \right)} \right| = - \dfrac{\alpha }{{2m}}\left( 0 \right) + C
Simplifying the terms, we get:
lnA(0)=0+C\Rightarrow \ln \left| {A\left( 0 \right)} \right| = 0 + C
lnA(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:
lnA=α2mt+C\ln \left| A \right| = - \dfrac{\alpha }{{2m}}t + C
lnA=α2mt+lnA(0)\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:
lnAlnA(0)=α2mt\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:
ln(AA(0))=α2mt\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 lnN=x\ln N = x , hence we get N=exN = {e^x} , i.e., lnN=xN=ex\ln N = x \Rightarrow N = {e^x} hence, applying this to the equation 4 we get:
ln(AA(0))=α2mt\ln \left( {\dfrac{A}{{A\left( 0 \right)}}} \right) = - \dfrac{\alpha }{{2m}}t
AA(0)=eα2mt\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:
A=A(0)eα2mt\Rightarrow A = A\left( 0 \right){e^{ - \dfrac{\alpha }{{2m}}t}}
So, the correct answer is “ A=A(0)eα2mt 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.