Quiz 2 (3-25-2009) solutions

Quiz 2 (3-25-2009) solutions - AMS 361: Applied Calculus IV...

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Unformatted text preview: AMS 361: Applied Calculus IV by Prof. Y. Deng Quiz 2 Wednesday (03/25/2009) at 8:05AM-9:25AM Open Book and Open Own Lecture Notes; Calculators Allowed (1) Do Any Two of the Three Problems; Mark the Two Problems You Attempt; Will Only Grade the First Two if You Don’t Mark the Problems You Attempted (2) Each Problem Is Worth 7.5 Points (3) No Points for Guessing Work and for Solutions without Appropriate Intermediate Steps; Partial Credits are given only for Steps that are Relevant to the Solutions Q2-1 (7.5 Points): Find the general solution of the following non-homogeneous equation Solution: The homogeneous characteristic equation is cos 1 , , => 1 , 4 The particular solution 1 1 1 , 4 = cos Therefore, the general solution is sin where Q2-2 (7.5 Points): For a given linear DE: √ 1 and are constants, find the general solution with the following two conditions: (3.0 Points) and (4.5 Points). Solution: The homogeneous general solution is IF Plug = cos suppose the particular solution: , , to , suppose the particular sol. IF Plug , to sin . sin and find , then 1 So, the general solution cos sin 1 . and find , then find 1 2 cos So, the general solution 0, sin 1 . and Q2-3 (7.5 Points): For a 2nd order linear DE: 0 where and are given functions of independent variable . You are given a solution , for this equation as (1) (5.0 Points) Find the general solution in terms of , , and . Solution: 0 : the given solution. By reduction of order method, Let’s determine is a second linearly independent solution of (*). which in (*). By substitution, 0 2 is a separable equation because we know We can get . by integration. After we get , then is another solution. . Therefore, the general solution is 2 (2) (2.5 Points) Find the general solution of 1 . given 2 0 with Solution: Let x = vt , x ' = tv '+ v, x ' ' = tv ' '+ v '+ v ' = v ' '+2v ' . Then the equation is 1 1 2 v' ' 4t 2 − 2 = = − − . 2 v' t (1 − t ) 1 − t 1 + t t ln v ' = −2 ln t − ln(t + 1) − ln(t − 1) + ln A. v" (t − t 3 ) + v' (2 − 4t 2 ) = 0. 1 A 1 1 1 (t + 1) 2 A v' = 2 = −{ 2 − + } A, v = A ln + + B. 1 t (1 + t )(t − 1) t 2(t + 1) 2(t − 1) t 2 (t − 1) 1 Let A = 1, B = 0, we have an independent solution x = tv = t ln (t + 1) 2 (t − 1) Therefore, the general solution is x = c1t ln (t + 1) (t − 1) 2 1 2 1 2 . + c1 + c2t. 1 2 . + 1. ...
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This note was uploaded on 05/11/2009 for the course AMS 361 taught by Professor Staff during the Spring '08 term at SUNY Stony Brook.

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