Lecture 4.9: Variation of parameters for systems Matthew Macauley - PowerPoint PPT Presentation

Lecture 4.9: Variation of parameters for systems Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 2080, Differential Equations M. Macauley (Clemson) Lecture 4.9: Variation of parameters for systems Differential Equations 1 / 6

Variation of parameters for non-systems Variation of parameters is a “last resort” method for finding a particular solution, y p ( t ). First order ODE: y ′ + p ( t ) y = f ( t ) � p ( t ) dt . (i) Solve y ′ h + p ( t ) y h = 0: get y h ( t ) = Cy 1 ( t ) = Ce − (ii) Find a particular solution of the form � � � p ( t ) dt p ( t ) dt dt . y p ( t ) = v ( t ) y 1 ( t ) = e − f ( t ) e Second order ODE: y ′′ + p ( t ) y ′ + q ( t ) y = f ( t ) (i) Solve y ′′ h + p ( t ) y ′ h + q ( t ) y h = 0: get y h ( t ) = C 1 y 1 ( t ) + C 2 y 2 ( t ). (ii) Find a particular solution of the form y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) . It turns out that � � − y 2 ( t ) f ( t ) dt y 1 ( t ) f ( t ) dt v 1 ( t ) = v 2 ( t ) = y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) , y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) . These methods always work , assuming that you can find y h ( t ), and evaluate the integrals. M. Macauley (Clemson) Lecture 4.9: Variation of parameters for systems Differential Equations 2 / 6

Variation of parameters for systems 2 × 2 system: x ′ ( t ) = A ( t ) x ( t ) + b ( t ) (i) Solve x ′ h = Ax h : get � x 11 ( t ) � � x 21 ( t ) � � x 11 ( t ) � � C 1 � x 21 ( t ) x h ( t ) = C 1 x 1 ( t ) + C 2 x 2 ( t ) = C 1 + C 2 = x 12 ( t ) x 22 ( t ) x 12 ( t ) x 22 ( t ) C 2 . � �� � X h ( t ) C (ii) Find a particular solution of the form x p ( t ) = X h ( t ) v ( t ): � x 11 ( t ) � � x 21 ( t ) � � x 11 ( t ) �� v 1 ( t ) � x 21 ( t ) x p ( t ) = v 1 ( t ) x 1 ( t )+ v 2 ( t ) x 2 ( t ) = v 1 ( t ) + v 2 ( t ) = x 12 ( t ) x 22 ( t ) x 12 ( t ) x 22 ( t ) v 2 ( t ) . � �� � X h ( t ) v ( t ) M. Macauley (Clemson) Lecture 4.9: Variation of parameters for systems Differential Equations 3 / 6

A specific example Example 1 � 1 � � 10 cos t � � 10 � − 4 x ′ = Solve the initial value problem x + , x (0) = . 2 e − t 2 − 5 4 M. Macauley (Clemson) Lecture 4.9: Variation of parameters for systems Differential Equations 4 / 6

A general example Example 2 Solve y ′′ + p ( t ) y ′ + q ( t ) y = f ( t ) by turning it into a 2 × 2 system first. M. Macauley (Clemson) Lecture 4.9: Variation of parameters for systems Differential Equations 5 / 6

Summary The variation of parameters method finds a particular solution of an ODE, of the form: (i) y p ( t ) = v ( t ) y 1 ( t ) (1st order) (ii) y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) (2nd order) (iii) x p ( t ) = X h ( t ) v ( t ) ( n × n system). � X − 1 We saw here that x p ( t ) = X h ( t ) v ( t ) = X h ( t ) ( t ) b ( t ) dt . h A second order ODE y ′′ + p ( t ) y ′ + q ( t ) y = f ( t ) can be written as a system by setting x 1 = y , x 2 = y ′ , to get � 0 � x ′ � � � � x 1 � � 0 1 1 = + x ′ − q ( t ) − p ( t ) x 2 f ( t ) . 2 Remarks This is the only way we know how to find a particular solution of a non-automonous ODE, e.g., x ′ = A x + b ( t ). This method also works if A ( t ) is non-constant, assuming that we can actually find x h ( t ) = C 1 x 1 ( t ) + C 2 x 2 ( t ). Such a solution is only defined where the Wronskian � x 11 ( t ) � � y 1 ( t ) � x 21 ( t ) y 2 ( t ) W [ x 1 ( t ) , x 2 ( t )] := det W [ y 1 ( t ) , y 2 ( t )] := det or x 12 ( t ) x 22 ( t ) y ′ 1 ( t ) y ′ 2 ( t ) , is non-zero. M. Macauley (Clemson) Lecture 4.9: Variation of parameters for systems Differential Equations 6 / 6