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!
How do you do the algebra?
Thanks!
1 Answer
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) whent=0
Then:
ln|A(0)| = 0 + C
Leading to the Particular Solution:
ln|A| = -alpha/(2m)t + ln|A(0)|
And assuming that
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