A population $x(t)$ grows in time according to $\frac{dx}{dt}=(x-1)(2x-1).$ Knowing that $x(0)=0,$ after how much time does it reach reach $50\%$ of its ultimate value as time passes?
1. Integrate $\int (6x-4)^2 dx.$
2. Solve $\frac{dy}{dx}=\cos x$ to find $y(x),$ given that $y(0)=1.$
3. Evaluate $\lim\limits_{x \to -\infty} \big(\frac{3x^2+7x+2}{x^2+1}\big).$

How could you manipulate this differential equation into quantities you can integrate?

How can you rewrite the product of the resulting two fractions as a sum/difference?

What does “Ultimate value” mean in terms of time?

… how about letting time go to infinity?

Which limit to infinity do you need to evaluate to find the ultimate value of the population?

Separating the variables ($x$ on one side, $t$ on the other), we can rewrite the differential equation as $dt = \frac{dx}{(x-1)(2x-1)}.$ Use partial fractions to split the product: $dt=\big(\frac{1}{x-1} - \frac{2}{2x-1}\big)dx.$ We can now integrate (directly or otherwise):

\begin{aligned} \int{dt} &= \int{\frac{dx}{x-1}} - 2\int{\frac{dx}{2x-1}} \\ t &= \ln|x-1| - \ln|2x-1| + C \\ &= \ln \left| \frac{x-1}{2x-1} \right| + C \end{aligned}

Using the initial condition $x(0)=0,$ we find that that $C=0.$ Therefore $t=\ln \big|\frac{x-1}{2x-1}\big|,$ which we can rearrange to get $x(t)=\frac{1}{2} - \frac{1}{4e^t - 2}.$

Evaluating $\lim\limits_{t\to\infty} x(t)$ gives us the ultimate population. As $t\to$ approaches $\infty,$ $\frac{1}{4e^t - 2}$ approaches $0,$ so the ultimate population is $\frac{1}{2},$ and $50\%$ of this is $\frac{1}{4}.$ Substituting, we find that $t_{50\%}=\ln\frac{3}{2}.$