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: USA Posts: 2,370 Thanks: 1274 
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: 1,769 Thanks: 626 Math Focus: Yet to find out.  

Tags 
integration 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
recursive integration,integration done, how to get formula  Lazar  Calculus  1  December 28th, 2014 12:40 PM 
integration  M^RT  Calculus  1  August 7th, 2013 08:44 AM 
integration  anco995  Calculus  2  January 9th, 2013 06:41 AM 
Integration with sin  jsmith613  Calculus  5  November 22nd, 2010 01:34 PM 
Integration  MathematicallyObtuse  Calculus  8  November 3rd, 2010 01:26 AM 