Population Growth and Ultimate Value

Question

I was given this question and am unsure how to continue:

Question

A population $x(t)$ grows in time according to $\frac{\mathrm{d}x}{\mathrm{d}t} = (x-1)(2x-1)$. Knowing that $x(0)=0$, after how much time does it reach $50\%$ of its ultimate value as time passes?

Attempt

I first converted into partial fractions: $$\mathrm{d}t = \dfrac{\mathrm{d}x}{(x-1)(2x-1)} \\ \mathrm{d}t = \mathrm{d}x\left(\dfrac{1}{x-1} - \dfrac{2}{2x-1}\right)$$

Then integrated to get $$t = \ln\left|x-1\right| - \ln\left|2x-1\right| + C$$ Subbing in $x=0$ with $t=0$ (to determine $C=0$), then rearranging, I got: $$t = \ln\left|\dfrac{x-1}{2x-1}\right|$$

My next instinct was to determine the maximum time, but after differentiating $t$ with respect to $x$, I get this, which has no solutions for $\dfrac{\mathrm{d}t}{\mathrm{d}x} = 0$: $$\dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{1}{(2x-1)(x-1)} = \dfrac{1}{x-1} - \dfrac{2}{2x-1}$$

As you can see, I really just went in a circle. I don't know how to continue. Many thanks.

Answer 1

$$t = \ln\left|\dfrac{x-1}{2x-1}\right|$$

Looks good to me.

My next instinct was to determine the maximum time,

Why? Or more importantly, why should there be a “maximum time”? The problem states:

after how much time does it reach $50\%$ of its ultimate value as time passes?

So think instead about finding $x_\infty = \lim_{t\to\infty} x(t)$, and then finding $t^*$ such that $x(t^*) = \frac{1}{2} x_\infty$.

Answer 2

write your equation in the form $$\frac{dx}{(x-1)(2x-1)}=dt$$ and calculate the partial decomposition: $$\frac{1}{(x-1)(2x-1)}=\frac{A}{x-1}+\frac{B}{2x-1}$$