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Practice Paper 2 Question 19

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?

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Warm-up Questions

  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 \(\lim\limits_{x \to -\infty} \big(\frac{3x^2+7x+2}{x^2+1}\big).\)


  • Hint 1
    How could you manipulate this differential equation into quantities you can integrate?
  • Hint 2
    How can you rewrite the product of the resulting two fractions as a sum/difference?
  • Hint 3
    What does "Ultimate value" mean in terms of time?
  • Hint 4
    ... how about letting time go to infinity?
  • Hint 5
    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):

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

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 \(t_{50\%}=\ln\frac{3}{2}.\)

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