Science:Math Exam Resources/Courses/MATH103/April 2014/Question 05 (a)

MATH103 April 2014

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Question 05 (a)

A mutation occurs within a bacterial species that allows it to acquire antibiotic resistance. Through natural selection the abundance of the mutant trait increases in the population. Let $y$ be the fraction (i. e. $0\leq y\leq 1$) of the population, which carries the antibiotic resistance trait. Suppose $y$ satisfies the following differential equation: ${\frac {dy}{dt}}=(1-y)^{3},$ where $t$ is time (years). Determine all steady state solutions (or stationary solutions or equilibria) of the differential equation. Work must be shown for full marks. Simplify fully.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
A steady state solution $y$ satisfies ${\frac {dy}{dt}}=0$.

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. A steady state solution is a solution $y$ of the differential equation, in this case ${\frac {dy}{dt}}=(1-y)^{3}$, such that ${\frac {dy}{dt}}=0$. Putting together the right-hand sides of these two equations yields $(1-y)^{3}=0$. Thus, $y=1$.

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