How to solve $y'(t) = a y(t) \cdot (b - y(t))$?

Question

I have given following differential equation:

$$y'(t) = a y(t) \cdot (b - y(t))$$

where $y(0) = y_0$ and $a,b,y_0 > 0$. How can I solve this differential equation?

I have tried to separate the variables and ended up with $$ \int a ~dt = \int \frac{1}{y(t)(b - y(t))}~dy(y). $$ I don't know how to go further, because I don't know how to solve the right-side integral. I would appreciate if you could help me solve the integral or solve the equation with another approach.

Note: I have the methods of a first year physics student available.

Answer 1

You wish to solve $$at+c=\int\frac{1}{y(b-y)}~dy=\frac1b\int\left(\frac1y+\frac1{b-y}\right)~dy=\frac1b(\log y-\log(b-y))$$ Then $$\left(\frac{y}{b-y}\right)=Ke^{abt}$$ or equivalently $$y(t)=\frac{Kbe^{abt}}{1+Ke^{abt}}$$ Now use the initial condition $y(0)=y_0$ in the equation $\left(\frac{y}{b-y}\right)=Ke^{abt}$ to evaluate $K$. Then $$K=\frac{y_0}{b-y_0}$$ So $$y(t)=\frac{y_0be^{abt}}{b-y_0+y_0e^{abt}}$$

Answer 2

An alternative method, motivated by the comment by @HansEngler is to do a substitution. $u=\frac1y$. Then $$u'(t)=-\frac1{y^2}y'(t)=-\frac1{y^2}(ay(b-y))=\frac{ay^2-aby}{y^2}=a-abu(t)$$$$\begin{align}u'+abu&=a\\e^{abt}u'+abe^{abt}u&=ae^{abt}\\(e^{abt}u)'&=ae^{abt}\\e^{abt}u&=\frac1{bK}+\frac1be^{abt}\\u&=\frac1{bK}e^{-abt}+\frac1b=\frac{e^{-abt}+K}{Kb}\\y&=\frac{Kb}{e^{-abt}+K}=\frac{Kbe^{abt}}{1+Ke^{abt}}\end{align}$$ Which yields the same result.

Answer 3

$$\int\frac{dy}{y(b-y)}=-\frac1b\int\left(\frac1{y-b}-\frac1y\right)dy=-\frac1b\log\frac{y-b}y...\text{and etc.}$$