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 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: USA Posts: 2,529 Thanks: 1389 $y^\prime(t) = k y(t)(M - y(t))$ $\dfrac{dy}{dt}= k y(t)(M - y(t))$ $\dfrac{dy}{y(t)(M-y(t))}= k ~dt$ $\dfrac{\log (y)}{M}-\dfrac{\log (y-M)}{M} = k t + C$ $\log\left(\dfrac{y(t)}{y(t)-M}\right) = k M t + C$ (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. Thanks from MathAboveMeth Last edited by skipjack; December 21st, 2016 at 02:55 PM. December 17th, 2016, 09:26 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2202 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. Thanks from MathAboveMeth December 17th, 2016, 09:43 AM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 In case you don't "see" it, 1/[y(M-y)]=A/y + B/(M-y), and solve for A,B (partial fractions). December 21st, 2016, 05:09 AM   #5
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 Originally Posted by romsek $y^\prime(t) = k y(t)(M - y(t))$ $\dfrac{dy}{dt}= k y(t)(M - y(t))$ $\dfrac{dy}{y(t)(M-y(t))}= k ~dt$ $\dfrac{\log (y)}{M}-\dfrac{\log (y-M)}{M} = k t + C$ $\log\left(\dfrac{y(t)}{y(t)-M}\right) = k M t + C$ (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.
I found the following
y(t) = -e^(ktM)M/(1-e^(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: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra It's wrong because you have no constant of integration. December 21st, 2016, 06:12 AM   #7
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 Originally Posted by v8archie It's wrong because you have no constant of integration.
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: 20,919 Thanks: 2202 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
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 Originally Posted by MathAboveMeth 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.
\displaystyle \begin{align*} \mathrm{e}^{k\,t\,M + C} &= \mathrm{e}^{k\,t\,M}\,\mathrm{e}^C \\ &= A\,\mathrm{e}^{k\,t\,M} \end{align*}

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 Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Njprince94 Differential Equations 7 December 9th, 2015 06:20 AM philm Differential Equations 5 May 6th, 2015 10:41 AM engininja Calculus 2 February 22nd, 2011 01:06 PM engininja Calculus 4 September 22nd, 2010 11:58 PM zaserov Calculus 1 October 25th, 2007 01:11 PM

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