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

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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 (b)
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. Find the general solution to the differential equation above.

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
To solve this differential equation, use separation of variables: Move all the y's to the left hand side of the equation, move all the t's to the right hand side of the equation, and integrate the left hand side in y and the right hand side in t.

Checking a solution serves two purposes: helping you if, after having used the hint, 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

Separating variables and integrating as an indefinite integral, ${\begin{aligned}{\frac {y\;dy}{y^{2}-k^{2}}}&=-\lambda t\;dt\\\int {\frac {y\;dy}{y^{2}-k^{2}}}&=-\lambda \int t\;dt\end{aligned}}$ We make a simple substitution in the y integral u= y2-k2, du = 2y dy to find ${\begin{aligned}\int {\frac {{\frac {1}{2}}du}{u}}&=-\lambda {\frac {1}{2}}t^{2}+C_{1}\\{\frac {1}{2}}\ln(|u|)&=-\lambda {\frac {1}{2}}t^{2}+C_{1}\\\ln(|y^{2}-k^{2}|)&=-\lambda t^{2}+C_{2},\end{aligned}}$ with C2=2C1 is the constant of integration. Evaluating this at t=0, we find $C_{2}=\ln(|y_{0}^{2}-k^{2}|)$, hence ${\begin{aligned}\ln(|y^{2}-k^{2}|)&=\ln(|y_{0}^{2}-k^{2}|)-\lambda t^{2}\\|y^{2}-k^{2}|&=|y_{0}^{2}-k^{2}|e^{-\lambda t^{2}}\\y(t)^{2}&=k^{2}\pm |y_{0}^{2}-k^{2}|e^{-\lambda t^{2}}\\y(t)&=\pm {\sqrt {k^{2}\pm |y_{0}^{2}-k^{2}|e^{-\lambda t^{2}}}}\end{aligned}}$ To choose branches $\pm$ we need more specific initial conditions.

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