The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field.

# 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?

The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.

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

## Hints

• 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?

## 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 . Please do not write to this address regarding general admissions or course queries.