Inverse Operators

# Inverse Operators - Differential Equations Second Quarter...

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Unformatted text preview: Differential Equations Second Quarter SY 0506 ____________________________________________________________________________________________________________ Silva, Alviso and Llacuna – Mathematics Department, MIT 143 Lesson 29 Inverse Operators &amp;on-Homogeneous Equations 3. Inverse Operator Consider the equation, F(D) y P = R(x) ––––––– (1) then, ) ( ) ( 1 x R D f y p = Where ) ( 1 D f = inverse operator Note: y P depends on the type of function represented by R(x). R(x) may be: 1. Exponential Function 2. Trigonometric Function 3. Polynomial Function 4. Composite Function 4.1 Exponential Shift 4.2 X Shift Properties of Inverse Operators: Let ( ) 1 F D and ( ) 1 G D be inverse operators. 1. ( ) 1 F D ( ) af x = ( ) ( ) 1 a f x F D 2. ( ) 1 F D ( ) ( ) + f x g x = ( ) ( ) ( ) ( ) + 1 1 f x g x F D F D 3. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = = 1 1 1 1 1 1 f x f x f x F D G D F D G D G D F D Differential Equations Second Quarter SY 0506 ____________________________________________________________________________________________________________ Silva, Alviso and Llacuna – Mathematics Department, MIT 144 3.1 R(x) is an Exponential Function 3.1.1 Given: f(D) y P = e ax ( ) ( ) ax ax p e a f e D f y 1 1 = = Such that f ( a ) ≠ if f(a)=0, and f(D) contains the factor (D-a) n times, that is, f(D) = Ø(D)(D-a) n , then 3.1.2 ax p n e y a D D = − ) )( ( φ ax n p e a D D y ) )( ( 1 − = φ ) ( , ) ( ! ≠ = a a n e x ax n φ φ Example: Find the particular solution of the differential equation. 1. 5 ) 2 ( 2 = − − p y D D SOLUTION: [ ] 2 1 5 ) 2 ( 2 2 − − = − − D D y D D p ; 2 1 5 2 = − − = a e D D y x p − − = x p e y 2 2 1 5 − = 2 1 5 p y 2 5 − = p y 2. ( ) x p e y D D 2 2 2 4 4 = + + SOLUTION: Differential Equations Second Quarter SY 0506 ____________________________________________________________________________________________________________ Silva, Alviso and Llacuna – Mathematics Department, MIT 145 [ ( ) x p e y D D 2 2 2 4 4 = + + ] 4 4 1 2 + + D D + + = x p e D D y 2 2 2 4 4 1 + + = x p e D D y 2 2 4 4 1 2 ; a = 2 + + = x p e y 2 2 4 ) 2 ( 4 2 1 2 x p e y 2 8 1 = Seatwork: Find the particular solution of the differential equation. 1. (D 3 – 2D 2 – 5D + 6) y p = (e 2x + 3) 2 ans: y p = 1/18 (e 4x ) – 3/2 (e 2x ) + 3/2 2. (D 3 – 5D 2 + 8D – 4) y p = e 2x + 2e x + 3e –x ans: y p = x 2 /2 (e 2x ) + 2xe x – 1/6 (e-x ) Homework: Find the particular solution of the differential equation....
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## Inverse Operators - Differential Equations Second Quarter...

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