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Unformatted text preview: Math 20D Practice Exam I Solutions 1. (a) (10 points) Find the general solution to the differential equation t 3 y + 4 t 2 y = e t . Rewrite as y + 4 t y = e t t 3 . This is a linear firstorder equation, so next compute the integrating factor = e R (4 /t ) dt = e 4ln t = t 4 . Multiply through by t 4 to get t 4 y + 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 )....
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This note was uploaded on 06/12/2008 for the course MATH 20D taught by Professor Mohanty during the Spring '06 term at UCSD.
 Spring '06
 Mohanty
 Math

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