Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE

# Technique of variation of parameters main theorem for

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Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Motivating Example 1 Motivating Example: Consider the nonhomogeneous problem y 00 + 4 y = 3 csc( t ) , which is inappropriate for the Method of Undetermined Coefficients The homogeneous solution is y c ( t ) = c 1 cos(2 t ) + c 2 sin(2 t ) Generalize this solution to the form y ( t ) = u 1 ( t ) cos(2 t ) + u 2 ( t ) sin(2 t ) , where the functions u 1 and u 2 are to be determined Differentiate y 0 ( t ) = - 2 u 1 ( t ) sin(2 t ) + 2 u 2 ( t ) cos(2 t ) + u 0 1 ( t ) cos(2 t ) + u 0 2 ( t ) sin(2 t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (13/27)

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Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Motivating Example 2 Motivating Example: The general solution has the form y ( t ) = u 1 ( t ) cos(2 t ) + u 2 ( t ) sin(2 t ) Since there is one general solution, there must be a condition relating u 1 and u 2 The computations are simplified by taking the relationship u 0 1 ( t ) cos(2 t ) + u 0 2 ( t ) sin(2 t ) = 0 This simplifies the derivative of the general solution to y 0 ( t ) = - 2 u 1 ( t ) sin(2 t ) + 2 u 2 ( t ) cos(2 t ) Differentiating again yields: y 00 ( t ) = - 4 u 1 ( t ) cos(2 t ) - 4 u 2 ( t ) sin(2 t ) - 2 u 0 1 ( t ) sin(2 t )+2 u 0 2 ( t ) cos(2 t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (14/27)
Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Motivating Example 3 Motivating Example: The differential equation is y 00 + 4 y = 3 csc( t ) , so substituting the general solution gives - 4 u 1 ( t ) cos(2 t ) - 4 u 2 ( t ) sin(2 t ) - 2 u 0 1 ( t ) sin(2 t ) +2 u 0 2 ( t ) cos(2 t ) + 4( u 1 ( t ) cos(2 t ) + u 2 ( t ) sin(2 t )) = 3 csc( t ) , which simplifies to - 2 u 0 1 ( t ) sin(2 t ) + 2 u 0 2 ( t ) cos(2 t ) = 3 csc( t ) This equation is combined with our earlier simplifying condition u 0 1 ( t ) cos(2 t ) + u 0 2 ( t ) sin(2 t ) = 0 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (15/27)

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Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Motivating Example 4 Motivating Example: The previous equations give two linear algebraic equations in u 0 1 and u 0 2 u 0 1 ( t ) cos(2 t ) + u 0 2 ( t ) sin(2 t ) = 0 - 2 u 0 1 ( t ) sin(2 t ) + 2 u 0 2 ( t ) cos(2 t ) = 3 csc( t ) The first equation gives u 0 2 ( t ) = - u 0 1 ( t ) cot(2 t ) It follows that (with trig identities) u 0 1 ( t ) = - 3 2 csc( t ) sin(2 t ) = - 3 cos( t ) and (with trig identities) u 0 2 ( t ) = 3 cos( t ) cot(2 t ) = 3 2 csc( t ) - 3 sin( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (16/27)
Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Motivating Example 5 Motivating Example: We solve the equations for u 0 1 and u 0 2 u 0 1 ( t ) = - 3 cos( t ) , so u 1 ( t ) = - 3 sin( t ) + c 1 Simlarly, u 0 2 ( t ) = 3 2 csc( t ) - 3 sin( t ) , so u 2 ( t ) = 3 2 ln | csc( t ) - cot( t ) | + 3 cos( t ) + c 2 It follows that the general solution is y ( t ) = u 1 ( t ) cos(2 t ) + u 2 ( t ) sin(2 t ) = - 3 sin( t ) cos(2 t ) + 3 2 sin(2 t ) ln | csc( t ) - cot( t ) | + 3 cos( t ) sin(2 t ) + c 1 cos(2 t ) + c 2 sin(2

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