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

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 (c)
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. Suppose y0 = 3k. Write the solution to the differential equation above in the form $y(t)=\cdots .$ What happens as $t\to \infty$?

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
Plug in $y_{0}=3k$ into the solution to Problem 4b. Remember that the square root must be positive, and that k (and therefore $y_{0}$) is also positive, and use this to pick the appropriate signs in the expression. Then take the limit as $t\to \infty$.

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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. Substituting y02=(3k)2, we find ${\begin{aligned}y(t)&=\pm {\sqrt {k^{2}\pm |(3k)^{2}-k^{2}|e^{-\lambda t^{2}}}}\\&=\pm {\sqrt {k^{2}\pm 8k^{2}e^{-\lambda t^{2}}}}\end{aligned}}$ When t=0, the argument to the square root is $k^{2}\pm 8k^{2}$, which must be positive. This indicates that we need to take the positive branch inside the square root: $y(t)=\pm {\sqrt {k^{2}+8k^{2}e^{-\lambda t^{2}}}},$ and because y(0) > 0, we need to take the positive branch outside the square root, giving the final answer ${\begin{aligned}y(t)&={\sqrt {k^{2}+8k^{2}e^{-\lambda t^{2}}}}\\&=|k|{\sqrt {1+8e^{-\lambda t^{2}}}}\\&=k{\sqrt {1+8e^{-\lambda t^{2}}}}\end{aligned}}$ where the last line holds as k is positive. As $t\to \infty$ we have that $y(t)\to +k.$

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