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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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    @Jorda$n$: If you have$n$'t see$n$ 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