How do you solve the Initial value probelm $dp/dt = 10p(1-p),    p(0)=0.1$? The problem is...
$$ \frac{dp}{dt} = 10p(1-p),$$   $p(0)=0.1$.
Solve and show that $p(t) \to 1$ as $t\to \infty.$
I know this is probably really simple, I was trying to go down the line of finding a general solution and then imposing the boundary condition.  But I can't even see how to find the general solution...
Then for the second bit taking the limit of the general solution $p(t)$?
 A: This is the so-called logistic equation, which occurs often in population dynamics and many other contexts. There's a trick which works for this particular equation and is much simpler than separation of variables (in my opinion): change variables to $y(t)=1/p(t)$. Then the nonlinear equation for $p$ turns into an inhomogeneous linear equation for $y$, which can be solved immediately by the usual "homogeneous + particular solution" method (the homogeneous solution is an exponential, and the particular solution is a constant). Since this is tagged as homework, I'll let you have a go at the details yourself.
A: HINT: The method we can use here is called Separation of Variables. Take all the $p$'s to one side and $t$'s to the other, then integrate both sides like this: $$ \int \frac{dp}{10p(1-p)} = \int dt $$ and now integrate both sides. The right hand side is simply $ t+ C$ where C is some constant. To integrate the left hand side, use partial fractions. Set $$ \frac{1}{10p(1-p) } = \frac{A}{p} + \frac{B}{1-p} $$ and solve for $A $ and $B$, then both terms are easily integrated in terms of natural logs. After that, you will be able to solve for $p$ in terms of $t$ and some constant. Feel free to ask for more help if you need it!
A: From your comment, it looks you have been able to integrate correctly, following Ragib's Hint and Gourtaur comment. But now your problem is (to finish the solution) to express $p(t)$. This rest part is a simple algebra. Let me express $p(t)$ in terms of $t$:
$\frac{p}{1-p}=e^{10t+10c}=e^{10t}.e^{10c}=k.e^{10t}$ (where $k=e^{10c}$ is a new constant)
$\Rightarrow \frac{p}{(1-p)+p}=\frac{ke^{10t}}{1+ke^{10t}}$ (I applied $\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a}{b+a}=\frac{c}{d+c}$. You can just multiply both sides by $(1-p)$, or cross-multiply and solve for $p$)
$\Rightarrow p=p(t)=\frac{1}{1+k'e^{-10t}}$ (dividing numerator and denominator of the fraction on RHS by $ke^{10t}$ and writing $k'=\frac{1}{k}$)
Now, from using the condition $p(0)=0.1=\frac{1}{10}$, we get $\frac{1}{10}=\frac{1}{1+k'}\Rightarrow k'=9$
Hence, you get $p(t)$ and when $t\to\infty$, $e^{-10t}$ tends to what?...
