# 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?

## Related topics

## Warm-up Questions

- Integrate \(\int (6x-4)^2 dx.\)
- Solve \(\frac{dy}{dx}=\cos x\) to find \(y(x),\) given that \(y(0)=1.\)
- Evaluate \(\lim\limits_{x \to -\infty} \big(\frac{3x^2+7x+2}{x^2+1}\big).\)

## Hints

- Hint 1How could you manipulate this differential equation into quantities you can integrate?
- Hint 2How can you rewrite the product of the resulting two fractions as a sum/difference?
- Hint 3What does "Ultimate value" mean in terms of time?
- Hint 4... how about letting time go to infinity?
- Hint 5Which limit to infinity do you need to evaluate to find the ultimate value of the population?

## Solution

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

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at oi.[email protected]. Please do not write to this address regarding general admissions or course queries.