MATH103 April 2012
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Question 04 (d)
Consider the differential equation
where is a positive constant, t ≥ 0, k > 0, but y may be positive or negative. Suppose
y(0) = y0.
Suppose y0 = k/2. Write the solution to the differential equation above in the form
What happens as ?
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Plug in into the solution of Problem 4b. Remember that and use this to select the appropriate signs in the expression. Then take the limit as .
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Using similar reasoning as the previous problem,
When t=0, y(0)>0, so we need to take the positive branch outside the square root:
again the last equality holding since k is positive. If we evaluate this at t=0, the positive branch gives , which is wrong, since y(0)=y0=k/2. Hence this is the wrong branch. Indeed, the negative branch gives Our answer is therefore
As we have that
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