Science:Math Exam Resources/Courses/MATH103/April 2010/Question 01 (h)

MATH103 April 2010

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Question 01 (h)
Multiple Choice Question: Exactly ONE of the answers provided is correct. There is no partial credit in this questions. Consider the differential equation ${\frac {dy}{dt}}=1-4y^{2}$ and initial condition $y(0)=0.$ After a long time (when $y$ no longer changes), the value of $y$ is i. $y=0$ ii. $y=1$ iii. $y=1/4$ iv. $y=4$ v. $y=1/2$

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
When $y(t)$ no longer changes, what does that imply for ${\frac {dy}{dt}}$?

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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. Saying that y(t) does not change any more is the same as saying that y(t) is constant. A function is constant if its derivative ${\frac {dy}{dt}}$ is equal to 0. So we look for values of $y$ that satisfy $\ 0=1-4y^{2}$. This equation only holds for $y=\pm 1/2$. Hence the the right answer is v. (Note: The answer y=-1/2 is not a choice here because of the initial condition y(0)=0. The solution of the differential equation starts at y=0 and then increases to y=1/2. With a different initial condition the other equilibrium y=-1/2 would be approached.)

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