How to do the ordinary differential equation? {dA}/{dt}=dotA=-alpha/{2m}A rarr A(t)=A(0)e^{-alpha/{2m}t}

{dA}/{dt}=dotA=-alpha/{2m}A rarr A(t)=A(0)e^{-alpha/{2m}t}

How do you do the algebra?
Thanks!

1 Answer
Jan 3, 2018

Given that:

dot A =-alpha/(2m)A

Then we can write:

(dA)/(dt) = -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 \ 1/A \ dA = int \ -alpha/(2m) \ dt

Which is directly integrable, to give:

ln|A| = -alpha/(2m)t + C

If we use the suggested initial condition:

A=A(0) when t=0

Then:

ln|A(0)| = 0 + C

Leading to the Particular Solution:

ln|A| = -alpha/(2m)t + ln|A(0)|

And assuming that A is positive over its range, then:

lnA - lnA(0) = -alpha/(2m)t

:. ln (A/(A(0))) = -alpha/(2m)t

:. A/(A(0)) = e^(-alpha/(2m)t)

And finally:

A = A(0)e^(-alpha/(2m)t) \ \ \ \ QED