Solving $\frac{dP}{dt} = k(M - P)$ 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.
 A: 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 $M$.
A: 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. 
