Chapter 4: Higher-Order Differential Equations Part 3 Department - PowerPoint PPT Presentation

Solving Systems of Linear Differential Equations Summary Chapter 4: Higher-Order Differential Equations – Part 3 Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw October 24, 2013 1 / 17 DE Lecture 7 王奕翔 王奕翔

Solving Systems of Linear Differential Equations Summary 1 Solving Systems of Linear Differential Equations 2 Summary 2 / 17 DE Lecture 7 王奕翔

Solving Systems of Linear Differential Equations Summary 3 / 17 see how to solve it So far we focus on solving an ODE with only one dependent variable. Systems of Linear Differential Equations DE Lecture 7 ⇒ 1 independent variable. Ordinary Differential Equation = ⇒ > 1 independent variables. Partial Differential Equation = Here we look at a system of linear ODE ( > 1 dependent variables), and Example : Let t be the independent variable, x , y be dependent variables. { x ′′ + 2 x ′ + y ′′ = x + 3 y + sin t . x ′ + y ′ = − 4 x + 2 y + e − t 王奕翔

Solving Systems of Linear Differential Equations Let us use the differential operator to rewrite it as follows: 4 / 17 Idea : Summary = DE Lecture 7 How to Solve? { x ′′ + 2 x ′ + y ′′ = x + 3 y + sin t . x ′ + y ′ = − 4 x + 2 y + e − t { D 2 { x } + 2 D { x } + D 2 { y } = x + 3 y + sin t D { x } + D { y } = − 4 x + 2 y + e − t {( ) ( ) D 2 + 2 D − 1 D 2 − 3 { x } + { y } = sin t ⇒ ( D + 4) { x } + ( D − 2) { y } = e − t Eliminate y and get a linear DE of x to solve x ( t ) . Eliminate x and get a linear DE of y to solve y ( t ) . 王奕翔

Solving Systems of Linear Differential Equations (1b) 5 / 17 To eliminate x : Summary To eliminate y : Elimination (1a) DE Lecture 7 { L 11 { x } + L 12 { y } = g 1 ( t ) L 21 { x } + L 22 { y } = g 2 ( t ) where L 11 = D 2 + 2 D − 1 , L 12 = D 2 − 3 , L 21 = D + 4 , L 22 = D − 2 . (1a) × L 22 − (1b) × L 12 = ⇒ ( L 11 L 22 − L 21 L 12 ) { x } = L 22 { g 1 }− L 12 { g 2 } (1a) × L 21 − (1b) × L 11 = ⇒ ( L 12 L 21 − L 22 L 11 ) { x } = L 21 { g 1 }− L 11 { g 2 } 王奕翔

Solving Systems of Linear Differential Equations (2a) 6 / 17 (1) and find out all additional constraints. ̸ Summary (2b) DE Lecture 7 (1a) Similar to the Cramer’s Rule, after the elimination we get (1b) { L 11 { x } + L 12 { y } = g 1 ( t ) L 21 { x } + L 22 { y } = g 2 ( t ) { L { x } = � g 1 ( t ) L { y } = � g 2 ( t ) � � � � � � � � � � � � g 1 ( t ) g 1 ( t ) � L 11 L 12 � � L 12 � � L 11 � where L = � g 1 ( t ) = � g 2 ( t ) = � , � , � � � � . g 2 ( t ) g 2 ( t ) L 21 L 22 L 22 L 21 ⇒ Solutions of (1) = ⇐ = Solutions of (2) Note : We should always plug the solutions found in solving (2) back to 王奕翔

Solving Systems of Linear Differential Equations . . . . . . . . . . . Summary . . . . . . . . . L kk j -th column replaced by g k 4 Plug into the initial system, find additional constraints on the coefficients 7 / 17 . DE Lecture 7 Solving a System of Linear DE with Constant Coefficients 1 Convert it into the following form:  L 11 { y 1 } + L 12 { y 2 } + · · · + L 1 k { y k } = g 1 ( t )     L 21 { y 1 } + L 22 { y 2 } + · · · + L 2 k { y k } = g 2 ( t )     L k 1 { y 1 } + L k 2 { y 2 } + · · · + L kk { y k } = g k ( t ) 2 Use Cramer’s rule to get L { y j } = � g j ( t ) , j = 1 , . . . , k , where � � � � · · · L 11 L 12 L 1 k � � � � · · · � L 21 L 22 L 2 k � L = � g j ( t ) = L | � � [ ] T � � , · · · g 1 � � � � · · · L k 1 L k 2 3 Solve each y j ( t ) , j = 1 , . . . , k . in the complimentary solutions { y 1 c , y 2 c , . . . , y kc } , and finalize. 王奕翔

Solving Systems of Linear Differential Equations Summary 8 / 17 find the solution of the other dependent variable (see example). plug the general solution we found back to the original system and after solving one dependent variable, it may save some time if we operators (see example). DE Lecture 7 additional constraints on the coefficients in the complimentary It is very important to plug the general solutions found for Notes and Tips { y 1 , y 2 , . . . , y k } back to the original system of equations to find solutions { y 1 c , y 2 c , . . . , y kc } (see example). When solving L { y j } = � g j ( t ) , sometimes we can eliminate redundant When k = 2 , that is, only two dependent variables to be solved, 王奕翔

Solving Systems of Linear Differential Equations (3a) 9 / 17 D Summary D Based on the method mentioned above, we compute (3b) DE Lecture 7 Solve Example A: Rewrite it as { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 {( ) D 2 − 4 { x } + D 2 { y } = t 2 ( D + 1) { x } + D { y } = 0 � � D 2 − 4 � � D 2 � = − ( D 2 + 4 D ) = − D ( D + 4) � � L = � D + 1 � � � � D 2 − 4 � � � � { t 2 } { t 2 } t 2 D 2 t 2 � � � � � g 1 ( t ) = � = D � g 2 ( t ) = � = − ( D + 1) , � � 0 D + 1 0 王奕翔

Solving Systems of Linear Differential Equations the the following: 10 / 17 Question : why can we cancel the repeated operators on both sides? D Summary DE Lecture 7 Hence, solutions of the original system of equations must be solutions of Example: Convert into two separate linear equations Solve { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 { ( ) { t 2 } { x } = � − � L { x } = � g 1 ( t ) ⇐ ⇒ D ( D + 4) { t 2 } L { y } = � g 2 ( t ) ⇐ ⇒ ( − D ( D + 4)) { y } = − ( D + 1) Ans : because instead of eliminating y by D { (3a) } − D 2 { (3b) } , we can simply eliminate y by (3a) − D { (3b) } . 王奕翔

Solving Systems of Linear Differential Equations (4a) 11 / 17 = Summary (4b) DE Lecture 7 Solve We convert the above into Solution 1: Solving x and y separately { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 { ( D + 4) { x } = − t 2 ( D ( D + 4)) { y } = t 2 + 2 t Step 1. Solve (5a): x c = c 1 e − 4 t , x p = A + Bt + Ct 2 , and − t 2 = ( D + 4) { x p } = (4 A + B ) + (4 B + 2 C ) t + 4 Ct 2 ⇒ C = − 1 4 , B = − 1 2 C = 1 8 , A = − 1 4 B = − 1 32 . Hence x ( t ) = c 1 e − 4 t − 1 32 + 1 8 t − 1 4 t 2 . 王奕翔

Solving Systems of Linear Differential Equations We convert the above into 12 / 17 Summary (5b) (5a) = DE Lecture 7 Solution 1: Solving x and y separately Solve { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 { ( D + 4) { x } = − t 2 ( D ( D + 4)) { y } = t 2 + 2 t Step 2. Solve (5b): y c = c 2 + c 3 e − 4 t , y p = At + Bt 2 + Ct 3 , and t 2 + 2 t = ( D 2 + 4 D ) { y p } = (4 A + 2 B ) + (8 B + 6 C ) t + 12 Ct 2 ⇒ C = 1 12 , B = 1 4 − 3 4 C = 3 16 , A = − 1 2 B = − 3 32 . Hence y ( t ) = c 2 + c 3 e − 4 t − 3 32 t + 3 16 t 2 + 1 12 t 3 . 王奕翔

Solving Systems of Linear Differential Equations Summary 13 / 17 DE Lecture 7 Solution 1: Solving x and y separately Solve { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 We find that for some c 1 , c 2 , c 3 , { c 1 e − 4 t − 32 + 1 1 8 t − 1 4 t 2 x ( t ) = c 2 + c 3 e − 4 t − 16 t 2 + 3 3 12 t 3 1 y ( t ) = 32 t + Final Step. Plug them back to find the constraints on { c 1 , c 2 , c 3 } . { (12 c 1 + 16 c 3 ) e − 4 t + t 2 = t 2 − (3 c 1 + 4 c 3 ) e − 4 t = 0 which implies c 3 = − 3/4 c 1 . Hence, the final solution is { c 1 e − 4 t − 32 + 1 1 8 t − 1 4 t 2 x ( t ) = 4 c 1 e − 4 t − 16 t 2 + c 2 − 3 3 3 12 t 3 1 y ( t ) = 32 t + 王奕翔

Solving Systems of Linear Differential Equations Plug this back to the second equation, we get 14 / 17 Plug it back to the first equation, it is also satisfied. Done! = Summary DE Lecture 7 Solution 2: Solving x first and then plugging in to find y Solve { x ′′ − 4 x + y ′′ = t 2 x ′ + x + y ′ = 0 After Step 1., we find that for some c 1 , x ( t ) = c 1 e − 4 t − 32 + 1 1 8 t − 1 4 t 2 . y ′ = − x ′ − x = 3 c 1 e − 4 t − 3 32 + 3 8 t + 1 4 t 2 ⇒ y = c 2 − 3 4 c 1 e − 4 t − 3 32 t + 3 16 t 2 + 1 12 t 3 王奕翔

Solving Systems of Linear Differential Equations Summary 1 Solving Systems of Linear Differential Equations 2 Summary 15 / 17 DE Lecture 7 王奕翔

Solving Systems of Linear Differential Equations Summary Short Recap Solving Systems of Linear DEs by Systematic Elimination 16 / 17 DE Lecture 7 代回原式以找出 complimentary solutions 中係數的關係 王奕翔

Solving Systems of Linear Differential Equations Summary Self-Practice Exercises 4-9: 1, 7, 11, 15, 19, 21 17 / 17 DE Lecture 7 王奕翔