t c 1 cos2 t c 2 sin2 t which shows the homogeneous and particular solutions

T c 1 cos2 t c 2 sin2 t which shows the homogeneous

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t ) + c 1 cos(2 t ) + c 2 sin(2 t ) , which shows the homogeneous and particular solutions Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations Part 2 - Nonhomogeneous — (17/27) Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Variation of Parameters 1 Technique of Variation of Parameters: Consider the nonhomogeneous problem y 00 + p ( t ) y 0 + q ( t ) y = g ( t ) , where p , q , and g are given continuous functions Assume we know the homogeneous solution : y c ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) Try a general solution of the form y ( t ) = u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t ) , where the functions u 1 and u 2 are to be determined Differentiating yields y 0 ( t ) = u 1 ( t ) y 0 1 ( t ) + u 2 ( t ) y 0 2 ( t ) + u 0 1 ( t ) y 1 ( t ) + u 0 2 ( t ) y 2 ( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (18/27) Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Variation of Parameters 2 Variation of Parameters: As before, there must be a condition relating u 1 and u 2 , so take u 0 1 ( t ) y 1 ( t ) + u 0 2 ( t ) y 2 ( t ) = 0 This simplifies the derivative of the general solution to y 0 ( t ) = u 1 ( t ) y 0 1 ( t ) + u 2 ( t ) y 0 2 ( t ) Differentiating again yields: y 00 ( t ) = u 1 ( t ) y 00 1 ( t ) + u 2 ( t ) y 00 2 ( t ) + u 0 1 ( t ) y 0 1 ( t ) + u 0 2 ( t ) y 0 2 ( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations Part 2 - Nonhomogeneous — (19/27) Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Variation of Parameters 3 Variation of Parameters: We now have expressions for the general solution , y ( t ) = u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t ), and its derivatives, y 0 ( t ) and y 00 ( t ), which we substitute into the nonhomogeneous problem: y 00 + p ( t ) y 0 + q ( t ) y = g ( t ) , This can be written in the form: u 1 ( t ) [ y 00 1 ( t ) + p ( t ) y 0 1 ( t ) + q ( t ) y 1 ( t )] + u 2 ( t ) [ y 00 2 ( t ) + p ( t ) y 0 2 ( t ) + q ( t ) y 2 ( t )] + u 0 1 ( t ) y 0 1 ( t ) + u 0 2 ( t ) y 0 2 ( t ) = g ( t ) The quantities in the square brackets are zero , since y 1 and y 2 are solutions of the homogeneous equation , leaving u 0 1 ( t ) y 0 1 ( t ) + u 0 2 ( t ) y 0 2 ( t ) = g ( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (20/27)

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Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Variation of Parameters 4 Variation of Parameters: This gives two linear algebraic equations in u 0 1 and u 0 2 u 0 1 ( t ) y 1 ( t ) + u 0 2 ( t ) y 2 ( t ) = 0 u 0 1 ( t ) y 0 1 ( t ) + u 0 2 ( t ) y 0 2 ( t ) = g ( t ) Recall Cramer’s Rule for solving a system of two linear equations in two unknowns, which above are the functions u 0 1 ( t ) and u 0 2 ( t ). u 0 1 ( t ) = det 0 y 2 ( t ) g ( t ) y 0 2 ( t ) det y 1 ( t ) y 2 ( t ) y 0 1 ( t ) y 0 2 ( t ) and u 0 2 ( t ) = det y 1 ( t ) 0 y 0 1 ( t ) g ( t ) det y 1 ( t ) y 2 ( t ) y 0 1 ( t ) y 0 2 ( t ) From before we recognize the denominator as the Wronskian : W [ y 1 , y 2 ]( t ) = det y 1 ( t ) y 2 ( t ) y 0 1 ( t ) y 0 2 ( t ) = y 1 ( t ) y 0 2 ( t ) - y 2 ( t ) y 0 1 ( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations Part 2 - Nonhomogeneous — (21/27) Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters
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