Homework 7 solutions

Homework 7 solutions - AMS 361 Applied Calculus IV by Prof...

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Unformatted text preview: AMS 361: Applied Calculus IV by Prof Y. Deng Homework 7 Assignment Date: Collection Date: Grade: Wednesday (03/25/2009) Wednesday (04/15/2009) Each problem is worth 10 points. Problem 7.1 Find the particular solution for sin tan Solution. We have two cases: 0. We take the particular solution is . (i). 0. Then (ii) 0. We have the general solution for the homogeneous solution cos sin . By the formula, the particular solution , where cos , sin , sin tan , W= . sin tan Therefore sin . We calculate sin sin 1 1 cos 2 2 sin sin 1 1 cos cos ln|cos | cos sin sin 1 cos 1 d(cos . cos ) d cos cos d cos ( ( Problem 7.2 Find the particular solution for 1 1 0, 1 Solution. The characteristic equation is 1, , , cos . Let Then , and return to the equation, we have 3 2 , 9 4 , 27 8 . Therefore we have 27 9 3 1,8 4 2 1, 1, , , , 1. The particular solution is 1 0. 1, and 1 1 40 Problem 7.3 Find the general solution for 1 15 1 4 1 1 Solution. Let , we have , 0 , + , and 2 2 1 +. For the homogeneous equation, we have the characteristic equation 2 2 1 0. By the general formula of roots in cubic function, Let 2 1, 9 2, 27 2, 47, 1, 2 9 , 27 d 4 , 3 2241 , then , 1 3 , 1 3 1 2 uv3 . 22 Then . We let 2 we have , then , 4 1/2, . , 1, 8 1/5 , and return the equation, 1/2 1. 1/5, The general solution is Problem 7.4 Find the general solution of the following equation: 1 sin Solution. We have two cases: 0. The characteristic equation 0. (i). 0. Then . The general solution is (ii). 0. Let . Then we have 0 is 0. The characteristic equation for formula, we have the general solution cos sin 1 sin sin 1 sin sin 0. 1 sin , . We let . . By the 1 sin ω 1 sin sin cos cos cos sin 1 sin 2 1 sin 2 1 [ cos cos 2 cos 85 2 2 . . Therefore sin 5 cos 4 2 22 4 cos 2 cos 4 sin 2 sin 1 1 4 4 3 6 6 sin 1 44 xcos 2 85 , and 5 2 sin 2 2 2 sin 2 8 8 3 6 5 2 cos sin 2 sin 3 4 4 1 6 We let 2 6 4 cos 6 2 cos 6 sin 2 3 2 2 2 sin . cos cos 2 sin sin , sin cos , . , and the general solution is Problem 7.5 Find the general solution of the following equation: 2 Solution. The characteristic equation is 2 0. 1 0. Then 1, 2, , , and We have 1 2 cos sin . , and return the equation, we have Let 5 4 3 2 , 20 12 +6Cx+2D, 60 24 6, 120 24 . Then we have 2 1, 2 5 0, 2 4 20 0, 2 3 12 60 0, 120 24 6 2 2 0, 6 24 2 2 0. Therefore , , , , , . We get . The general solution . 3 ...
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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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