November 1st, 2016, 03:20 PM  #1 
Senior Member Joined: Feb 2015 From: london Posts: 121 Thanks: 0  Integration
$\displaystyle \frac{dm}{dt} = (yu) m$ I am trying to solve the above differential equation using the initial condition m(0) = a $\displaystyle \frac{dm}{dt} \frac{1}{m} = (yu) $ $\displaystyle ln(m) = (yu)t $ $\displaystyle m = e^{(yu)t} + C $ Using initial condition we find c = a, therefore $\displaystyle m = e^{(yu)t} + a $ However answer in my solutions is the below, so not sure where I have gone wrong? $\displaystyle m = ae^{(yu)t} $ 
November 1st, 2016, 03:31 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,264 Thanks: 650 
after integrating you get $\ln(m) = (yu)t + \ln(a)$ where $\ln(a)$ is the constant of integration now take the exponential of both sides $m = ae^{(yu)t}$ just a little trick since the constant is free to be anything 
November 1st, 2016, 06:53 PM  #3 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 
Why has this been posted in PreCalculus when Integration/DEs are clearly Calculus???

November 1st, 2016, 07:42 PM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,224 Thanks: 420 Math Focus: Yet to find out.  

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