Science:Math Exam Resources/Courses/MATH103/April 2012/Question 04 (f)

MATH103 April 2012

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q2 (f)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q4 (e)  •  Q4 (f)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  

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Question 04 (f)
Consider the differential equation ${\frac {dy}{dt}}=-\lambda t\left({\frac {y^{2}-k^{2}}{y}}\right),$ where $\lambda$ is a positive constant, t ≥ 0, k > 0, but y may be positive or negative. Suppose y(0) = y0. Sketch your solutions, including those from part (a).

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
We want to graph the solutions to the differential equation with initial conditions $y(0)=3k$, $y(0)={\frac {k}{2}}$, and $y(0)=-{\frac {k}{2}}$. We know the limiting behavior of each of these cases. Use the differential equation to see where each solution is increasing and decreasing, and the fact that the solutions approach the constant solutions calculated in part a, to sketch the graphs.

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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. First, we graph the constant solutions to the differential equation: $y=k$ and $y=-k.$ These are the solutions we got from part a. Then, we graph the solution from part (c). Since 3k is larger than the constant solution k, it follows that the graph must stay above this line (since if y = k at any point $t_{0}$ , then y must be equal to k for all $t>t_{0}$). Furthermore, if this holds, then y^2 > k^2, so from the differential equation given in the problem statement, ${\frac {dy}{dt}}$ is negative and therefore the function is decreasing. Now, we graph the solutions from part (d) and (e). Since ${\frac {k}{2}}$ and $-{\frac {k}{2}}$ are between -k and k, it follows that the graphs for each of these solutions must lie between the constant solutions y=-k and y=k. The graph from part d has $y(0)={\frac {k}{2}}$. Note that for any $0<y<{\frac {k}{2}}$, dy/dt is positive, so the solution to the differential equation given in (d) should be increasing. We know that the limit of the function as $t\to \infty$ is k. Similarly, if we plug a y-value such that -k < y < 0 into the differential equation, we see that the solution should be decreasing, so the graph of the solution described in part e should decrease to its limit -k.

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