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November 1st, 2016, 04:20 PM   #1
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Integration

$\displaystyle \frac{dm}{dt} = (y-u) m$

I am trying to solve the above differential equation using the initial condition m(0) = a

$\displaystyle \frac{dm}{dt} \frac{1}{m} = (y-u) $

$\displaystyle ln(m) = (y-u)t $

$\displaystyle m = e^{(y-u)t} + C $

Using initial condition we find c = a, therefore

$\displaystyle m = e^{(y-u)t} + a $

However answer in my solutions is the below, so not sure where I have gone wrong?
$\displaystyle m = ae^{(y-u)t} $
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November 1st, 2016, 04:31 PM   #2
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after integrating you get

$\ln(m) = (y-u)t + \ln(a)$

where $\ln(a)$ is the constant of integration

now take the exponential of both sides

$m = ae^{(y-u)t}$

just a little trick since the constant is free to be anything
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November 1st, 2016, 07:53 PM   #3
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Why has this been posted in Pre-Calculus when Integration/DEs are clearly Calculus???
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November 1st, 2016, 08:42 PM   #4
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Originally Posted by Prove It View Post
Why has this been posted in Pre-Calculus when Integration/DEs are clearly Calculus???
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