Science:Math Exam Resources/Courses/MATH103/April 2016/Question 06 (a) (iv)

MATH103 April 2016

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Question 06 (a) (iv)
Solve the differential equation in part (ii) for $y(t)$ with the initial condition $y(0)$ found in part (ii). For full credit you must show your work.

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.

Hint 1
Note that the differential equation in part (ii) is separable.

Hint 2
Consider using partial fractions.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. We can rewrite the differential equation in the following separable form: ${\frac {1}{y(1-y)}}\,dy={\frac {1}{2}}\,dt,\ y(0)=0.2.$ Integrating both sides, we get $(*)\quad \int {\frac {dy}{y(1-y)}}=\int {\frac {dt}{2}}.$ We first compute the right hand side: $\int {\frac {dt}{2}}={\frac {t}{2}}+C_{1}.$ The left hand side can be computed using partial fractions: $\int {\frac {dy}{y(1-y)}}=\int \left({\frac {1}{y}}+{\frac {1}{1-y}}\right)dy=\ln(y)-\ln(1-y)+C_{2}=\ln \left({\frac {y}{1-y}}\right)+C_{2}.$ (Note: Since the population $y$ will never be negative nor exceed the carrying capacity, $y,1-y\geq 0$, so the absolute value signs in the antiderivatives can be omitted.) Returning to $(*),$ we have $\ln \left({\frac {y}{1-y}}\right)={\frac {t}{2}}+C_{3},\quad C_{3}:=C_{1}-C_{2}.$ Thus ${\frac {y}{1-y}}=e^{{\frac {t}{2}}+C_{3}}=e^{\frac {t}{2}}e^{C_{3}}=Ce^{\frac {t}{2}},\quad C:=e^{C_{3}}.$ We now substitute in the initial value $y(0)=0.2.$ So, ${\frac {0.2}{1-0.2}}={\frac {1}{4}}=Ce^{\frac {0}{2}}=C.$ We finally have ${\frac {y}{1-y}}={\frac {1}{4}}e^{\frac {t}{2}},$ which yields $\color {blue}y(t)={\frac {{\frac {1}{4}}e^{\frac {t}{2}}}{1+{\frac {1}{4}}e^{\frac {t}{2}}}}.$