285lect26 - 1 Lecture 27 Example 1 (Continue) Z Z Z d (sin...

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1 Lecture 27 Example 1 (Continue) u 2 ( x ) = Z cos x csc 2 xdx = Z cos x sin 2 x xdx = Z d (sin x ) sin 2 x = & 1 sin x : IV. Particular solution: y p ( x ) = ln & & & cot x 2 & & & ± cos x + ± & 1 sin x ² sin x = ln & & & cot x 2 & & & ± cos x & 1 : General solution: y ( x ) = y p ( x ) + y c ( x ) = ln & & & cot x 2 & & & ± cos x & 1 + C 1 cos x + C 2 sin x: Example 2 y 00 & 2 y 0 & 8 y = 3 e & 2 x I. Complimentary equation: y 00 & 2 y 0 & 8 y = 0 : Characteristic equation: Complimentary solution: y c ( x ) = C 1 e & 2 x + C 2 e 4 x : II. Particular solution: y p ( x ) = u 1 ( x ) e & 2 x + u 2 ( x ) e 4 x : 1

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III. Equations for u 0 1 and u 0 2 : u 0 1 e & 2 x + u 0 2 e 4 x = 0 & 2 u 0 1 e & 2 x + 4 u 0 2 e 4 x = 3 e & 2 x ) u 0 1 e & 2 x + u 0 2 e 4 x = 0 & u 0 1 e & 2 x + 2 u 0 2 e 4 x = 3 2 e & 2 x II + I ) 3 u 0 2 e 4 x = 3 2 e & 2 x ) u 0 2 = 1 2 e & 6 x : u 0 1 e & 2 x + u 0 2 e 4 x = 0 ; u 0 2 = 1 2 e & 6 x ) u 0 1 e & 2 x + e 4 x 2 e & 6 x = 0 ) u 0 1 e & 2 x + e & 2 x 2 = 0 ) u 0 1 = & 1 2 : Thus, we have u 1 ( x ) = & Z 1 2 dx = & x 2 : u 2 ( x ) = 1 2 Z e & 6 x dx = & 1 12 e & 6 x : IV. Particular solution: y p ( x ) = & x 2 e & 2 x & 1 12 e & 6 x e 4 x = & 1 12 e & 2 x (6 x + 1) : V. General solution: y ( x ) = y p ( x ) + + y c ( x ) = & 1 12 e & 2 x (6 x + 1) + C 1 e & 2 x + C 2 e 4 x : Example 3 (For independent study) y 00 & 4 y = sinh 2 x: I. Complimentary equation: y 00 & 4 y = 0 : 2
Characteristic equation: r 2 & 4 = 0 : Complimentary solution: y c ( x ) = C 1 e 2 x + C 2 e & 2 x = A cosh 2 x + B sinh 2 x: Particular solution. Method I: u 0 1 cosh 2 x + u 0 2 sinh 2 x = 0 I 2 u 0 1 sinh 2 x + 2 u 0 2 cosh 2 x = sinh 2 x II

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This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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285lect26 - 1 Lecture 27 Example 1 (Continue) Z Z Z d (sin...

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