Homework-8-Solution-f2009

# Homework-8-Solution-f2009 - AMS 361 Applied Calculus IV by...

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Unformatted text preview: AMS 361: Applied Calculus IV by Prof Y. Deng Homework 8 Assignment Date: Collection Date: Grade: Wednesday (11/11/2009) Wednesday (11/18/2009) NO HW on 11/25 Each problem is worth 10 points. Problem 8.1: Define two operators Applying these two operators to a function , we may get and check if . What if we define the two operators as follows, check it again: If , Check if ( ( = c bc , . c ( c bc c 1 Because L1 and L2 are independent of t, so the operations are commutative, i.e. 1. If = ), ), check if . ))+ t D( t )=t 1 ))+t( t ))= 2 2t ) +t ) Because L1 and L2 are NOT independent of t, so the operations are NOT commutative, i.e. ≠ Problem 8.2: Solve the following system of equations with given initial conditions: 2 0 0 2: Plug 5 into 1 : y : 0 0 1 2 3 4 5 y y 10 y , 2y , 0 (5): C cos t C sin t (4): =0 y t C sin t C cos t + C sin t (3): =0 In fact, 0 0 2 Problem 8.3: Solve the following system of equations by (1) substitution method and (2) operation method: 2 6 1. Substitution Method： (2): 3 : 1: 20 0 2 (3) 3 4 5 0 20 0 5, 4. 3 1 2 Ce Ce 4C e 5C e 4C e 5C e )+ Ce Ce Ce Ce Ce Ce 1 Ce 2 Ce 2. Operation Method： 2 6 2 6 2 6 6 0 1 : 2 18 20 0 3 1 12 1 6 6 3 3 1 12 12 18 2 18 18 0 0 5, 20 4. 0 01 02 18 1 2 1 2 0 4 5 Similarly, we have: Ce 4C e Ce 5C e 3 4C e 5C e )+ Ce Ce Ce Ce Ce Ce 1 Ce 2 Ce Problem 8.4: Find the general solution of the following. 1 2 3 (1)+(3) we get: Let , 4 Integrating factor: : (5)* Integrate both sides, (2): Integrate both sides, yt (3): (6) Integrating factor: (5)* : Integrate both sides, t 1 (4) (5) (3): t t 1 4 t In fact, yt t 1 Alternatively, we can use the operation method: 1 1 1 1 1 1 1 1 =0 0 0 Calculate the determinate the operation matrix, we get: 0 : 0 0, 1, 1 So we can write our solution in the form of : A B C (*) Note that here we add a before because on the other side of the equation we have . To make our solution independent, need to add t here. So now we can plug (*) into the system of DE’s, and solve for the relationships of the constant coefficients. Then we obtain the same answer: t yt 1 2 t 2 1 5 Problem 8.5: Use operational determinant method to solve the following system of equations: 1 1 1 (2)*( (3)-(4): 1 1 e 5 2: y In fact, 1e 0 0 e 0 : 2: 1 2 1 2 1 1 2 2 = 2e 2 2e (5) 0 2 0 2 1 2 2e 2 0 (3) 4 2 2 2e 6 ...
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