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December 17th, 2016, 09:06 AM   #1
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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.
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December 17th, 2016, 09:13 AM   #2
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$y^\prime(t) = k y(t)(M - y(t))$

$\dfrac{dy}{dt}= k y(t)(M - y(t))$


$\dfrac{dy}{y(t)(M-y(t))}= k ~dt$

$\dfrac{\log (y)}{M}-\dfrac{\log (y-M)}{M} = k t + C$

$\log\left(\dfrac{y(t)}{y(t)-M}\right) = k M t + C$ (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.
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Last edited by skipjack; December 21st, 2016 at 02:55 PM.
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December 17th, 2016, 09:26 AM   #3
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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.
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December 17th, 2016, 09:43 AM   #4
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In case you don't "see" it,

1/[y(M-y)]=A/y + B/(M-y), and solve for A,B (partial fractions).
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December 21st, 2016, 05:09 AM   #5
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Quote:
Originally Posted by romsek View Post
$y^\prime(t) = k y(t)(M - y(t))$

$\dfrac{dy}{dt}= k y(t)(M - y(t))$


$\dfrac{dy}{y(t)(M-y(t))}= k ~dt$

$\dfrac{\log (y)}{M}-\dfrac{\log (y-M)}{M} = k t + C$

$\log\left(\dfrac{y(t)}{y(t)-M}\right) = k M t + C$ (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.
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December 21st, 2016, 05:41 AM   #6
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It's wrong because you have no constant of integration.
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December 21st, 2016, 06:12 AM   #7
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Quote:
Originally Posted by v8archie View Post
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.
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December 21st, 2016, 03:13 PM   #8
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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.
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December 22nd, 2016, 12:46 AM   #9
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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.
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December 22nd, 2016, 03:07 AM   #10
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Quote:
Originally Posted by MathAboveMeth View Post
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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