December 17th, 2016, 09:06 AM  #1 
Newbie Joined: Dec 2016 From: Netherlands Posts: 15 Thanks: 0  Solve differential equation with use of separation of variables
Hello, I'm not sure whether this is an actual calculus question, but I hope you can help me. y'(t)= k* y(t) *(M − y(t)). I need to solve this equation by separation of variables. I watched some videos on this, but these didn't really help. Anyone that can help? Last edited by skipjack; December 21st, 2016 at 02:56 PM. 
December 17th, 2016, 09:13 AM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,511 Thanks: 761 
$y^\prime(t) = k y(t)(M  y(t))$ $\dfrac{dy}{dt}= k y(t)(M  y(t))$ $\dfrac{dy}{y(t)(My(t))}= k ~dt$ $\dfrac{\log (y)}{M}\dfrac{\log (yM)}{M} = k t + C$ $\log\left(\dfrac{y(t)}{y(t)M}\right) = k M t + C$ (note i just set MC to C as it's an arbitrary constant) With a bit more algebra, you can find an expression for $y(t)$ I leave that to you. Last edited by skipjack; December 21st, 2016 at 02:55 PM. 
December 17th, 2016, 09:26 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,063 Thanks: 1396 
dt/dy = 1/(ky(M  y)) = 1/(kMy) + 1/(kM(M  y)) Note that y = 0 and y = M are solutions. Other solutions are valid on a restricted domain. 
December 17th, 2016, 09:43 AM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90 
In case you don't "see" it, 1/[y(My)]=A/y + B/(My), and solve for A,B (partial fractions). 
December 21st, 2016, 05:09 AM  #5  
Newbie Joined: Dec 2016 From: Netherlands Posts: 15 Thanks: 0  Quote:
y(t) = e^(ktM)M/(1e^(ktM) Just wanted to know if this is correct, because in another question I had to compare this result with another result and they weren't really alike. Last edited by skipjack; December 21st, 2016 at 02:54 PM.  
December 21st, 2016, 05:41 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra 
It's wrong because you have no constant of integration.

December 21st, 2016, 06:12 AM  #7 
Newbie Joined: Dec 2016 From: Netherlands Posts: 15 Thanks: 0  Oh oops I forgot that indeed; would the solution be right if you put +C*M in the power of e?
Last edited by skipjack; December 21st, 2016 at 02:52 PM. 
December 21st, 2016, 03:13 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,063 Thanks: 1396 
I would suggest replacing e^(ktM + CM) with Ce^(ktM) and allowing this new C to be zero or negative. That would compensate for not using $\ln\left\dfrac{y(t)}{y(t)M}\right$ earlier. 
December 22nd, 2016, 12:46 AM  #9 
Newbie Joined: Dec 2016 From: Netherlands Posts: 15 Thanks: 0 
But how would I write this down? I can't figure out how to get to Ce^ktM from ln(y(t)/(y(t)M)) = ktM +C.
Last edited by skipjack; December 22nd, 2016 at 02:57 AM. 
December 22nd, 2016, 03:07 AM  #10  
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35  Quote:
where $\displaystyle \begin{align*} A = \mathrm{e}^C \end{align*}$, as $\displaystyle \begin{align*} \mathrm{e}^C \end{align*}$ is also a constant.  

Tags 
differential, equation, separation, solve, variables 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Separation of variables  Njprince94  Differential Equations  7  December 9th, 2015 06:20 AM 
Using Separation of variables  philm  Differential Equations  5  May 6th, 2015 10:41 AM 
Separation of Variables  engininja  Calculus  2  February 22nd, 2011 01:06 PM 
Separation of Variables  engininja  Calculus  4  September 22nd, 2010 11:58 PM 
Separation of Variables  zaserov  Calculus  1  October 25th, 2007 01:11 PM 