Joseph M Mahaffy h mahaffymathsdsuedu i Lecture Notes Second Order Linear

# Joseph m mahaffy h mahaffymathsdsuedu i lecture notes

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Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (9/32) Introduction Theory for 2 nd Order DEs Linear Constant Coefficient DEs Existence and Uniqueness Linear Operators and Superposition Wronskian and Fundamental Set of Solutions Wronskian Wronskian: Consider the linear homogeneous 2 nd order DE L [ y ] = y 00 + p ( t ) y 0 + q ( t ) y = 0 . with p ( t ) and q ( t ) continuous on an interval I Let y 1 and y 2 be solutions satisfying L [ y i ] = 0 for i = 1 , 2 and define the Wronskian by W [ y 1 , y 2 ]( t ) = y 1 ( t ) y 2 ( t ) y 0 1 ( t ) y 0 2 ( t ) = y 1 ( t ) y 0 2 ( t ) - y 0 1 ( t ) y 2 ( t ) . If W [ y 1 , y 2 ]( t ) 6 = 0 on I , then the general solution of L [ y ] = 0 satisfies y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (10/32) Introduction Theory for 2 nd Order DEs Linear Constant Coefficient DEs Existence and Uniqueness Linear Operators and Superposition Wronskian and Fundamental Set of Solutions Fundamental Set of Solutions Theorem Let y 1 and y 2 be two solutions of y 00 + p ( t ) y 0 + q ( t ) y = 0 , and assume the Wronskian, W [ y 1 , y 2 ]( t ) 6 = 0 on I . Then y 1 and y 2 form a fundamental set of solutions , and the general solution is given by y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) . where c 1 and c 2 are arbitrary constants. If there are given initial conditions, y ( t 0 ) = y 0 and y 0 ( t 0 ) = y 1 for some t 0 I , then these conditions determine c 1 and c 2 uniquely. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (11/32) Introduction Theory for 2 nd Order DEs Linear Constant Coefficient DEs Homogeneous Equations Method of Undetermined Coefficients Forced Vibrations Homogeneous Equations 1 Homogeneous Equation: The general 2 nd order constant coefficient homogeneous differential equation is written: ay 00 + by 0 + cy = 0 This can be written as a system of 1 st order differential equations ˙ x = Ax = 0 1 - c/a - b/a x , where x = x 1 x 2 = y y 0 This has a the general solution x = c 1 y 1 y 0 1 + c 2 y 2 y 0 2 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (12/32) Subscribe to view the full document.

Introduction Theory for 2 nd Order DEs Linear Constant Coefficient DEs Homogeneous Equations Method of Undetermined Coefficients Forced Vibrations Homogeneous Equations 2 Characteristic Equation: Obtain characteristic equation by solving det | A - λ I | = - λ 1 - c/a - b/a - λ = 1 a ( 2 + + c ) = 0 Find eigenvectors by solving ( A - λ I ) v = - λ 1 - c/a - b/a - λ v 1 v 2 = 0 0 If λ is an eigenvalue, then it follows the corresponding eigenvector is v = 1 λ Then a solution is given by x = e λt v = e λt λe λt = y ( t ) y 0 ( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (13/32) Introduction Theory for 2 nd Order DEs Linear Constant Coefficient DEs Homogeneous Equations Method of Undetermined Coefficients Forced Vibrations Homogeneous Equations 3 Theorem Let λ 1 and λ 2 be the roots of the characteristic equation 2 + + c = 0 .  • Fall '08
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