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Fall 2006 - Hall's Class - Practice Exam 1

Fall 2006 - Hall's Class - Practice Exam 1 - Math 20D...

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Math 20D Practice Exam I Solutions 1. (a) (10 points) Find the general solution to the differential equation t 3 y 0 + 4 t 2 y = e - t . Rewrite as y 0 + 4 t y = e - t t 3 . This is a linear first-order equation, so next compute the integrating factor μ = e R (4 /t ) dt = e 4 ln t = t 4 . Multiply through by t 4 to get t 4 y 0 + 4 t 3 y = te - t . The LHS is d dt ( t 4 y ), so taking the antiderivative of both sides gives t 4 y = - te - t - e - t + C . So the general solution is y ( t ) = - e - t t 3 - e - t t 4 + C t 4 . (b) (8 points) Find the particular solution to the differential equation in part (a) with initial value y (1) = 0. y (1) = - e - 1 - e - 1 + C = 0, so C = 2 e - 1 . The solution is y ( t ) = - e - t t 3 - e - t t 4 + 2 e - 1 t 4 . (c) (2 points) What is the largest interval on which the solution you found in part (b) is defined? The solution is defined on the interval (0 , ). 2. Consider the autonomous differential equation dy dt = f ( y ) , where f ( y ) = (9 - y 2 )(4 - y 2 ). (a) (10 points) Draw a phase line, and label each equilibrium solution as stable, unstable, or semistable.
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