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 December 17th, 2016, 10:06 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 03:56 PM.
 December 17th, 2016, 10:13 AM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,106 Thanks: 578 $y^\prime(t) = k y(t)(M - y(t))$ $\dfrac{dy}{dt}= k y(t)(M - y(t))$ $\dfrac{dy}{y(t)(M-y(t))}= k ~dt$ $\dfrac{\log (y)}{M}-\dfrac{\log (y-M)}{M} = k t + C$ $\log\left(\dfrac{y(t)}{y(t)-M}\right) = k M t + C$ (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 03:55 PM.
 December 17th, 2016, 10:26 AM #3 Global Moderator   Joined: Dec 2006 Posts: 16,775 Thanks: 1236 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, 10:43 AM #4 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 965 Thanks: 78 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, 06:09 AM   #5
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Quote:
 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 03:54 PM.

 December 21st, 2016, 06:41 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,540 Thanks: 2146 Math Focus: Mainly analysis and algebra It's wrong because you have no constant of integration.
December 21st, 2016, 07:12 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 03:52 PM.

 December 21st, 2016, 04:13 PM #8 Global Moderator   Joined: Dec 2006 Posts: 16,775 Thanks: 1236 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, 01:46 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 03:57 AM.
December 22nd, 2016, 04:07 AM   #10
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Quote:
 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.

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