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

MATH103 April 2014

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[hide] Question 05 (b)
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). At time $t=0$ it is determined that half of the bacterial population carries the antibiotic resistance trait. Determine the fraction of the bacterial population with antibiotic resistance as a function of time $t$ . 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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Use separation of variables to find the general solution to the differential equation.

[show] Hint 2
Use the initial condition to determine the constant in the general solution.

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[show] 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. Use the method of separation of variables. First, re-arrange the differential equation: ${\frac {dy}{dt}}=(1-y)^{3}\Rightarrow {\frac {\mathrm {d} y}{(1-y)^{3}}}=\mathrm {d} t.$ Next, integrate both sides. On the right-hand side, we have $\int \mathrm {d} t=t+C,$ where $C$ is a constant. On the left-hand side, we have $\int {\frac {\mathrm {d} y}{(1-y)^{3}}}.$ We can perform the integral, for instance, by substitution. Letting $u=1-y$ , so that $\mathrm {d} u=-\mathrm {d} y$ , or equivalently $\mathrm {d} y=-\mathrm {d} u$ , we get $\int {\frac {\mathrm {d} y}{(1-y)^{3}}}=\int {\frac {-\mathrm {d} u}{u^{3}}}={\frac {1}{2u^{2}}}={\frac {1}{2(1-y)^{2}}}.$ We have neglected to include the constant because it has already been taken into account above. Thus, we have found that ${\frac {1}{2(1-y)^{2}}}=t+C.$ To determine the constant, use the fact that $y(0)={\frac {1}{2}}$ . Plugging this into the equation above yields ${\frac {1}{2\left(1-{\frac {1}{2}}\right)^{2}}}=0+C,$ i. e. $2=C$ . Thus, ${\frac {1}{2(1-y)^{2}}}=t+2.$ Solving for $y$ gives $y=1-{\frac {1}{\sqrt {2t+4}}}.$ Note that we have taken the positive square root as we require the solution $y$ to lie in the interval $[0,1]$ .