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 November 1st, 2016, 03:20 PM #1 Senior Member   Joined: Feb 2015 From: london Posts: 101 Thanks: 0 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? $\displaystyle m = ae^{(y-u)t}$
 November 1st, 2016, 03:31 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 925 Thanks: 501 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 Thanks from topsquark
 November 1st, 2016, 06:53 PM #3 Member   Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 Why has this been posted in Pre-Calculus when Integration/DEs are clearly Calculus???
November 1st, 2016, 07:42 PM   #4
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 Originally Posted by Prove It Why has this been posted in Pre-Calculus when Integration/DEs are clearly Calculus???
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