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

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

How do you do the algebra?
Thanks!

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Answer 1 · 2018-01-03T15:20:15

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