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

MATH103 April 2012

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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.

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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. 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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