2_1a - Solutions 2.1-Page 106 Problem 3 A homogeneous...

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Solutions 2.1-Page 106 Problem 3 A homogeneous second-order linear differential equation, two functions and , and a pair of initial conditions are given. First verify that and are solutions of the differential equation. Then find a particular solution of the form 1 y 1 + 2 y 2 y 1 y 2 y 2 1 c y c y = that satisfies the given initial conditions. Primes denote derivatives with respect to x . 0 4 = + y y ; , x y 2 cos 1 = x y 2 sin 2 = ; 3 ) 0 ( = y , 8 ) 0 ( = y The problem will first be verified for , and then the problem will be verified again for . 1 y 2 y Differentiating twice yields: 1 y x y 2 cos 4 1 = Substituting into the differential equation yields: 0 ) 2 (cos 4 2 cos 4 = + x x Differentiating twice yields: 2 y x y 2 sin 4 2 = Substituting into the differential equation yields: 0 ) 2 (sin 4 2 sin 4 = + x x The general form of the particular solution is x c x c y 2 sin 2 cos 2 1 + = . Differentiating gives x c x c y 2 sin 2 2 cos 2 1 2 = .
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This note was uploaded on 06/06/2011 for the course EGM 3311 taught by Professor Haftka during the Spring '11 term at University of Florida.

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2_1a - Solutions 2.1-Page 106 Problem 3 A homogeneous...

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