quiz5 - Quiz 5 Name Math 240 Calculus III March 3 2009 Note...

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Unformatted text preview: Quiz 5 Name: Math 240 - Calculus III March 3, 2009 Note: In order to receive full credit, you must show work that justifies your answer. 1. Solve the following initial value problem for y ( t ): 4 t 2 y 00 + y = 0 , y (1) = 4 , y (1) = 0 Solution : We can convert the differential equation into one with constant coefficients by the change of variable t = e x . Then dy dt = dy dx 1 t and d 2 y dt 2 = d 2 y dx 2- dy dx 1 t 2 , and the equation becomes 4 d 2 y dx 2- dy dx + y = 4 d 2 y dx 2- 4 dy dx + y = 0 . The auxiliary equation is (2 m- 1) 2 = 0, which has a repeated root of 1 2 . The general solution is then y ( x ) = c 1 e x/ 2 + c 2 xe x/ 2 . But x = ln t , so y ( t ) = c 1 t 1 / 2 + c 2 t 1 / 2 ln t . The condition y (1) = 4 implies c 1 = 4, and the condition y (1) = 0 implies c 2 =- 2. Thus, the solution is y ( t ) = 4 t 1 / 2- 2 t 1 / 2 ln t. Note that we could also find the solution using the “book method” of assuming y = t m , from which we obtain 4 t 2 y 00 + y = [4 m ( m-...
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This note was uploaded on 09/09/2009 for the course MATH 240 taught by Professor Cremins during the Spring '07 term at Maryland.

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quiz5 - Quiz 5 Name Math 240 Calculus III March 3 2009 Note...

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