Winter 2006 - Reynold's Class - Quiz 2

Winter 2006 - Reynold's Class - Quiz 2 - Math 20D, Lecture...

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Math 20D, Lecture C, Winter 2006 27 January 2006 Quiz 2, version A Name: ID #: Section Time: Show all work clearly and in order, and circle your final answers. You have 20 minutes to take this 25 point quiz. 1. [ 5 points ] Determine, without solving the differential equation, an interval in which the solution of the given initial value problem is certain to exist. ( t 2 - 4 t - 12) y 0 + 3 y = 4 t, y ( - 3) = 2 Solution: We first convert to standard linear form, y 0 + 3 t 2 - 4 t - 12 y = 4 t 2 - 4 t - 12 The coefficient functions are discontinuous when the denominator equals zero, i.e. 0 = t 2 - 4 t - 12 = ( t - 6)( t + 2) . Thus the coefficient functions are discontinuous at the points t = - 2 and t = 6 . So according to Theorem 2.4.1 from the text, the differential equation is guaranteed to have a solution in the intervals -∞ < t < - 2 , - 2 < t < 6 , 6 < t < . Since our initial condition is specified at t = - 3 , the correct interval is -∞ < t < - 2 . 2.
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Winter 2006 - Reynold's Class - Quiz 2 - Math 20D, Lecture...

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