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MATH 2270 (28)
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Week 10.pdf

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Department
Mathematics
Course
MATH 2270
Professor
Matthew Demers
Semester
Fall

Description
1 6.4 Examples for Systems of First-Order Homogeneous Lin- ear Equations Now, we solve a few systems. Example: Find the general solution to the system ! x = 3 1 x: (6.55) 12 ▯1 First, we need to ▯nd the eigenvalues of the matrix of coe▯cients in this system. The eigenvalues are given by the characteristic equation: (3 ▯ ▯)(▯1 ▯ ▯) ▯ 12 = 0 ▯3 ▯ 2▯ + ▯ ▯ 12 = 0 ▯ ▯ 2▯ ▯ 15 = 0 (▯ ▯ 5)(▯ + 3) = 0: Thus, we obtain ▯1= 5 and ▯ 2 ▯3. Now, ▯nding eigenvectors, we ▯rst consider 1 , which was 5. Substituting this into the (A - I1 )x = 0, we obtain (in augmented matrix form): ! ▯2 1 0 (6.56) 12 ▯6 0 which reduces to ! ▯2 1 0 (6.57) 0 0 0 One row of zeroes gives us one free variable to work with, and thus one eigenvector. 1 Indeed, we ▯nd that if 2 = t, then ▯2x1+ t = 0, or 1 = 2t. Thus, we obtain the eigenvector ! 1 V1= : (6.58) 2 2 Now, consider the other eigenvalue2 ▯ = ▯3. Subbing into (A - 1▯ )x = 0 and reducing gives the following: ! 6 1 0 : (6.59) 0 0 0 We obtain an eigenvector of ! V = ▯1 : (6.60) 2 6 Therefore, our solution is ! ! ! y1 1 5t ▯1 ▯3t = C 1 e + C 2 e : (6.61) y2 2 6 Example 2: Find the general solution to the system ! ▯1 2 x = x: (6.62) ▯2 ▯2 The characteristic equation is (▯1 ▯ ▯)(▯2 ▯ ▯) + 4 = 0 2 2 + 3▯ + ▯ + 4 = 0 2 ▯ + 3▯ + 6 = 0 p 3 15 ▯ = ▯ ▯ j 2 2 Now we must ▯nd eigenvectors! We have a complex pair, so we’ll take the eigen- value corresponding to the \plus" case and work with it. Substituting this into the (A - I▯1)x = 0, and reducing, we get: p ! 1▯ 1j 2 0 2 2 : (6.63) 0 0 0 3 If we let2x = t, then p 1 15 (2 ▯ 2 j)x1+ 2t = 0; (6.64) or solving fo1 x , ▯4 x1= p t: (6.65) 1 ▯ 15j Put the imaginary component in the numerator, by multiplying the top and bot- tom by the conjugate of the denominator. We obtain p x = ▯4(1 + 15j) (6.66)
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