# Logistic equation solution

I have the following question:

The number $$x$$ in a population satisfies the logistic equation: $$\frac{dx}{dt} = 2x(10-x),$$ where $$t$$ is time in years. If the population is $$2$$ initially, find the time it takes to increase to $$4$$.

I used partial fractions to arrive at: $$(1/10)/x + (1/10)/(10-x) + C = 2t,$$ which leads me to: $$1/40*x^2 + x/2 + (x^2)/4 + C = x.$$ When $$t(0) = 2$$, then $$C$$ is $$2$$. I'm not sure where to go from here/am a bit unsure if this is even correct. Any help would be great,

Thanks :)

• This is a Separable Equation, so separate and integrate using $x(0)=2$ and you get $$x(t) = \dfrac{10 e^{20 t}}{e^{20 t}+4}$$ Clear? Now proceed and get $$t\approx 0.0490414626505863$$ – Moo Nov 7 '18 at 4:42

You have $$\frac{x'}{2x(10-x)}=1.$$ Partial fractions give you $$\frac{x'}{2x}+\frac{x'}{20-2x}=10.$$ Now, taking antiderivatives, $$\tfrac12\,\log x-\tfrac12\,\log(20-2x)=10t+c.$$ Rewrite as $$\log \frac{x}{20-2x}=20t+c.$$ From $$x(0)=2$$ we get $$c=\log\frac1{8}$$. Now exponentiating, $$\tag{*} \frac{x}{20-2x}=\exp(20t-\log 8)=\frac18\,e^{20t}.$$ Solving for $$x$$ we get $$8x=20e^{20t}-2xe^{20t}$$, so $$x(t)=\frac{20e^{20t}}{8+2e^{20t}}=\frac{10e^{20t}}{4+e^{20t}}.$$ And we want to find $$t$$ so that $$x(t)=4$$. We solve from $$(*)$$, $$e^{20t}=\frac{4x}{10-x}=\frac{16}{6}=\frac83,$$ so $$t=\frac1{20}\,\log\frac83\simeq 0.049041\ldots$$