Pre-Calc Homework Solutions 349

Pre-Calc Homework Solutions 349 - Section 8.4 x 3 2x 2 2 (x...

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Unformatted text preview: Section 8.4 x 3 2x 2 2 (x 2 1)2 Ax (x 2 B 1) Cx (x 2 D 1)2 1 dy y2 y 1 dy y(y 1) 1 y(y 1) 349 24. 27. (Cx C)x D) (B D) e x dx e x dx A y B y 1 x3 2x 2 2 (Ax Ax 3 B)(x 2 Bx 2 1) (A ex C Equating coefficients of like terms gives A 1, B 2, A C 0, B D 2 1 A(y (A 1) B)y B(y) A Solving the system simultaneously yields A 1, B 2, C 1, D 0. Equating coefficients of like terms gives A B 0 and A 1 1, B 1. x 3 2x 2 2 dx (x 2 1)2 x x2 x x2 1 1 ln (x 2 2 2 dx 1 x (x 2 2 x2 1 1)2 Solving the system simultaneously yields A 1 dx x (x 2 1)2 1 2(x 2 1) y(y 1) dy 1 dy y 1 y 1 dy C2 ln y dx C ln y ln 2 ln y ln y 0 1 1 0, y ex 2. ln y C 1 1 1 dx 1) dx 1 2 tan x Substitute x C or C 1 d ex ln 2 ln 2. 25. 1 1 +1 1 1 1 1 1 1 1 1 The solution to the initial value problem is ln y dy sin 28. 1 1 1 (y 1 1 0 d 0 d 1 0 0 d 1 0 ln 1 26. 2 1 1)2 1 dy sin d (y 1)2 1 cos C y 1 2 Substitute x 1 0 ,y 0. 1 ln 2 1 2 2 1 +1 1 1 1 2 2 C or C The solution to the initial value problem is 1 y 1 2 2 2 2 0 1 cos 1 cos 1 cos dx 3x 1 1 1 1 2 2 1 1 y 1 2 1 y d 0 d 0 1 2 d 29. dy 1 2 tan 0 1 0 x2 2 2 tan 1 2 x 2 3x 1 3x 2 2 (x A x 2 2)(x B x 1) 1 x2 1 1 A A(x (A B 1) B)x 0, A dx 3x B(x A 2) 2B Equating coefficients of like terms gives 2B 1 1, B 1. dx 2 x 2 x dx 1 Solving the system simultaneously yields A dy y ln x x2 2 ln x 3, y 0. 1 C Substitute x 0 y 0 ln x ln 2 2 C or C ln x 1 ln 2 ln 2 The solution to the initial value problem is ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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