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I am suppose to solve for P(t), to find an epxression for P(t) and I am suppose to find the limit.

I can't find anything.

$$\frac{dP}{dt} = k(M - P)$$

$$\frac{dP}{M - P} = k \, dt$$

$$\int \frac{dP}{M - P} = \int k \, dt$$ $$ \ln \frac{1}{M - P} = xk + c$$ $$ \frac{1}{M - P} = e^{xk} + e^c$$ $$ \frac{1}{e^{xk} + e^c} = M - P$$ $$ -\frac{1}{e^{xk} + e^c} +M= P$$

This is wrong but I am not sure why.

2 Answers 2

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The separation of variables went well, and in general outline the calculation was along the right lines. However, there are some problems of detail.

An antiderivative of $\frac{dP}{M-P}$ is $-\ln(|M-P|)$. In your work, the minus sign is missing.

It is always good to check by differentiating whether you have integrated right. The derivative of $\ln(M-P)$ with respect to $P$ is $-\frac{1}{M-P}$ (Chain Rule). Not quite the $\frac{1}{M-P}$ that is needed, but the fix is easy.

Later there is a typo, there is an $x$ where $t$ is intended. There is also a problem with the simplification of $e^{kt+c}$. Note that $e^{u+v}=e^u e^v$.

To do things right, we integrate and get $$-\ln(|M-P|)=kt +c.$$ Either multiply both sides by $-1$, and take the exponential of both sides, or exponentiate directly. We do the first. So we have $\ln(|M-P|)=-kt -c$, and therefore $|M-P|=e^{-c}e^{-kt}$, so $M-P=\pm e^{-c}e^{-kt}$.

For simplicity, let $C=\pm e^{-c}$. We then get $P=M-Ce^{-kt}.$ To find the appropriate value of $C$, we need more information, such as an initial condition, the value of $P$ at a certain time $t$, often (but not necessarily) at $t=0$. In particular, if $P(0)=0$, it turns out that $C=M$.

The limit as $t\to\infty$ is easy to find even if we are not given an initial condition. I assume that the constant $k$ is positive. Then, as $t\to\infty$, we have $e^{-kt}\to 0$, so the limit of $P(t)$ is $M$.

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    I still get the wrong answer fixing those mistakes though, I need $M - M^{-ekt}$2012-06-18
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    @Jordan: Think about what Andre cited. Reflect, you will find more than you wanted. :-)2012-06-18
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    @Jordan: I wrote out one solution. In order to get something like what you mention above, you will need the *initial condition* $P(0)=0$. Maybe that was a part of the problem that you didn't mention.2012-06-18
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    That still is not what the book gets and that is why I am confused.2012-06-18
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    @Jordan: If you tell us what the book gets, we can clarify things. As usual there are several possibilities, maybe I have a mistake, maybe the book does, or maybe, as has happened several times with questions that you ask, both answers are right but just *look* different. If there is an initial condition mentioned in the problem that you did not include, please make sure to mention them.2012-06-18
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    $P9t) = M - Me^{-kt}$2012-06-18
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    @Jordan: I checked in my *Stewart*, found the problem. He as assuming without saying so that at the beginning the amount $P(0)$ learned is $0$, which is very reasonable. So $P(0)=M-Ce^0=0$ and therefore $C=M$.2012-06-18
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    @AndréNicolas I do not follow why C = M.2012-06-18
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    @Jordan: This is a model of amount learned, presumably from knowing nothing at time $t=0$, though that is not explicit. Our general formula is $P(t)=M-Ce^{-kt}$. Put $t=0$. Then $e^{-kt}=e^0=1$, so $0=P(0)=M-C$. Since $M-C=0$, we have $C=M$.2012-06-18
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    @AndréNicolas So both answers are exactly the same? The book just chose an incredibly difficult method to show that?2012-06-18
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    Actually there is an even simpler way, let $y=M-P$. Then $\frac{dP}{dt}=-\frac{dy}{dt}$ and therefore $\frac{dy}{dt}=-ky$. this is the DE of exponential decay, $y=Ce^{-kt}$ and therefore $M-P=Ce^{-kt}$. I don't know what the solutions manual did. Probably something like what I wrote out in the answer.2012-06-18
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    @Jordan: As in another of your recent questions, you can also solve this ODE using other standard methods (integrating factor or hom.+part. solution). But the suggestion in André's latest comment is the fastest way.2012-06-19
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    @HansLundmark I have never heard of these standard methods, do you mean homogenous part and the part solution? I do no understand.2012-06-19
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    @Jordan: If you haven't seen those methods yet, you will most likely cover them later in your course. The method I'm talking about is when you find the general solution to the homogeneous equation with the help of the characteristic polynomial, and then find a particular solution which deals with the right-hand side of the original equation.2012-06-19
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I don't want to add anything important than André cited but, maybe mine illustrate the problem easier. :-)

$$\frac{dP}{dt} = k(M - P)$$

$$\frac{dP}{M - P} = k \, dt$$

$$\int \frac{dP}{M - P} = \int k \, dt$$ $$ \ln\left| \frac{1}{M - P}\right| = -kt + c$$ $$ \frac{1}{M - P} = e^{-kt+ c}=e^{-kt}.e^c=Ce^{-kt} $$ I think the rest is easy because you did the same above before in the body of question.

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    You need to say $\left|\dfrac{1}{M-P}\right| = Ce^{-kt}$ and $C$ is positive, and then drop the absolute value sign along with the assumption that $C>0$.2012-06-19