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 :)
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 $$ t=\frac1{20}\,\log\frac83\simeq 0.049041\ldots $$