# Initial Value Differential Equation

I have this problem that I don't know how to approach. I have tried it a lot of different ways, but it doesn't seem like any of them are the way my teacher wants me to do it.

$$\frac{\mbox{d}N}{\mbox{d}t} = 5N(3 - N)\ ,\ \ \ \ \ \ N(0) = 1$$ How does $N(t)$ behave as 't' approaches +infinity?

• What ways have you tried? – Wright-Moran Sep 18 '14 at 22:11
• I have learned separation, and for this equation that would leave me with dN/(5N(3-N))=dt, correct? – powanakaab Sep 18 '14 at 22:27

the general solution is this, easily determined by separation of variables and direct integration and some elementary algebra afterwards. (if you are interested in the steps, write me a comment and i'll add them) $$\mbox{N}(t)=\frac{3e^{15t}}{e^{15t}+c_1}, \mbox{with } c\in \mathbb{R}$$ forcing the initial condition: $$\mbox{N}(0)=\frac{3}{1+c_1}=1\in \mathbb{R}$$ solving for $c_1$ gives: $c_1=2$ so: the particular solution is: $$\mbox{N}(t)=\frac{3e^{15t}}{e^{15t}+2}$$ now let's find the limit: $$\lim_{t\to +\infty}\mbox{N}(t)=\lim_{t\to +\infty}\frac{3e^{15t}}{e^{15t}+2}=3$$
You have 2 straionary points $0$ and $3$. One of them is stable (asymptotically), one of them is not stable. Stability is easy to understand checking derivative around these point. So $0$ is obiously unstable since the derivative is positive when $N=\epsilon>0$ and negative when $N<0$. So the solution gets out from $0$-neighborhood. $3$ is the stable stationary point by the similar reasoning. So you solution since it does not start from zero will asymptotically converge to $3$.