Solutions of the differential equation $\frac{dx}{dt}=(1-x)x$

Question

In solving the differential equation $\frac{dx}{dt}=(1-x)x$, I can obtain one solution by integrating $\frac{1}{x(1-x)}dx $ with the assumption that $0<x<1$. How can I obtain other solutions?

Thanks in advance!

Answer 1

Writing $$\frac{dx}{(1-x)x}=dt$$ and integrating we get $$x(t)=\frac{e^t}{e^t+C}$$ Note that $$\frac{1}{(1-x)x}=\frac{1}{x}+\frac{1}{1-x}$$

Answer 2

Another approach..

$$x'={(1-x)x}$$ $$x'-x=-x^2$$ $$(xe^{-t})'=-(xe^{-t})^2e^t$$ $$ \int \frac {d(xe^{-t})}{(xe^{-t})^2}=-\int e^t dt$$ $$ (xe^{-t})^{-1}=e^t+K \implies x^{-1}=\frac {e^t+K}{e^t}$$ $$ \implies x=\frac {e^t} {e^t+K}$$

Answer 3

With $y:=1/x$, the equation linearizes

$$-\frac{y'}{y^2}=\left(1-\frac1y\right)\frac1y,$$

$$y'=1-y$$ and

$$y=1+Ce^{-t}=\frac1x.$$