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Unformatted text preview: Math 33B HW #7 Solutions Problems Graded Section 8.4: 28 Section 8.5: 8, 26 Section 9.1: 10, 20 Solutions Section 8.4: Problem 28 We follow the analysis in Example 4.8 (starting on page 356). We see that the first mass has the force from both springs acting on it while the second mass has the force from the second spring and the external force acting upon it. From Newton's second law, the equations are: Equation (1) cos Equation (2) Now, we let , , , , where we have: , , , We can now rewrite equations (1) and (2), respectively, as follows: Additionally, we have (from the definition of ): We can write these equations in matrix form: 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 Since both masses start from rest and at their equilibrium positions, we have the following initial conditions: 0 0 0, 0 0 0 In terms of our , we have: 0 0, 0, 0, 0 T and not simply , (i.e. we have an external force), the system Since we have is inhomogeneous. Section 8.5: Problem 8 The system in Problem 2 can be written as: 6 4 8 6 e and t We want to show that t e We first show x : e e 6 4 8 6 Similarly, we show y e 2e e e : 2e 4e 2e 2e 6e 8e 4e 6e e 2e are solutions to the system above. 2e 2e 6 4 4e e 6e 2e 8 6 2e 8e 12e 4e We now verify that any linear combination is also a solution. In other words, we want to show : 2 e 2 e 2e 2e 2 e 4 e 2e 4e 6 4 e e 8 6 e 2e e e 6 4 8 6 e 2 e e e 6 4 8 6 e 2 e 6 e e 4 e 2 e e e 2 e 8 6 e 6e 4e 6e 8e 6e 8e 12e 8e 2 e 2 e 2 e 4 e Section 8.5: Problem 26 We are given the following solutions: cos sin , cos The system and the initial conditions are: 1 1 1 , 0 0 2 1 We first verify that cos cos 1 2 cos cos cos 1 1 cos cos sin sin sin sin t cos cos sin and sin sin sin cos are solutions to the homogeneous solution: sin sin cos sin sin 1 2 sin sin sin 1 1 sin cos cos cos cos sin sin cos cos cos sin sin cos t We can show that the solutions are linearly independent by assuring that the Wronskian is not identically zero: , det cos sin sin cos cos cos cos sin sin cos sin sin t sin t sin sin sin cos cos t cos cos We see that the Wronskian is not identically zero and thus the solutions are linearly independent. Another, perhaps simpler, way to see if two solutions is by looking at a particular value of . Take an arbitrary , say 0, and see if one solution is a multiple of the other: cos 0 sin 0 0 cos 0 1 sin 0 cos 0 0 sin 0 0 Since the two solutions are not multiples of one another at one time instant, they are linearly independent at this one time instant. From Definition 5.13, if a set of solutions is linearly independent at one value of , then they are linearly independent for any value of . Therefore, the two solutions are linearly independent. Finally, we can get a general solution using the two given solutions: cos sin sin cos sin we use the initial condition to solve for the undermined coefficients: 1 0 0 cos 0 sin 0 sin 0 cos 0 0 cos 0 sin 0 1 cos 0 We have the following set of equations: 0 1 2 The solution is thus: sin 0 2 2 cos sin sin 2sin cos Section 9.1: Problem 10 We have the following: 1 0 0 4 3 2 8 4 3 Therefore, we have: 1 0 0 4 3 2 8 4 3 We want to find out when the determinant of the above matrix is zero in order to get the eigenvalues. It is simplest to take the determinant along the first row since it has two zeros: det 1 3 3 2 4 1 9 8 1 1 1 1 1 0 The eigenvalues are: 1 (multiplicity 2) 1 Section 9.1: Problem 20 We are given: 3 2 4 3 To get the eigenvalues: 3 2 4 3 det 3 3 2 4 9 8 1 1 The eigenvalues are: 1, 1 Now, we can get the eigenvectors for each eigenvalue. For 1: 3 2 1 4 3 The two equations are: 3 2 2 2 0 3 4 4 0 4 We see that the two equations lead to the same result of . Each equation resulting from the eigenvalue problem will always be multiples of the other equations. This shows that you can always just take one equation resulting from and know that the other equations do not 1 and thus our eigenvector is: give any new information. We arbitrarily set 1 1 For 1: 3 2 1 4 3 The two equations are: 3 2 4 2 0 3 4 2 0 4 Again, the two equations lead to the same result of 2 consequently 2 2. The eigenvector is: 1 2 The solution can then be written as: 1 1 . We arbitrarily set 1 and 1 2 ...
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 Spring '07
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