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Unformatted text preview: MA341 Homework 13 Solutions 9.5 31. x ( t ) = " 1 3 3 1 # x ( t ), x (0) = " 3 1 # . The characteristic equation for this matrix is A rI = 1 r 3 3 1 r = r 2 2 r 8 = ( r + 2)( r 4) = 0 . Hence, the eigenvalues of this matrix are r 1 = 2 and r 2 = 4. To find the eigenvectors that correspond to r 1 = 2, we must solve ( A r 1 I ) u = 0: " 3 3 3 3 #" u 1 u 2 # = " # . The solutions is u 1 = s , u 2 = s , where s is arbitrary. Therefore, the eigenvectors associated with r 1 = 2 are u 1 = s " 1 1 # . Taking s = 1, we have u 1 = " 1 1 # . For r 2 = 4, the equation ( A r 2 I ) u = 0 reads: " 3 3 3 3 #" u 1 u 2 # = " # . Its solution is u 1 = s , u 2 = s , where s is arbitrary. Therefore, the eigenvectors associated with r 2 = 4 are u 2 = s " 1 1 # . Taking s = 1, we obtain u 2 = " 1 1 # . Since the eigenvectors u 1 and u 2 are linearly independent, the general solution is x ( t ) = c 1 e 2 t " 1 1 # + c 2 e 4 t " 1 1 # = " e 2 t e 4 t e 2 t e 4 t #" c 1 c 2 # . To satisfy the initial condition, we solve x (0) = " 1 1 1 1 #" c 1 c 2 # = " 3 1 # . and find that c 1 = 1, c 2 = 2. Consequently, the solution to the initial value problem is x ( t ) = " e 2 t e 4 t e 2 t e 4 t #" 1 2 # = " e 2 t + 2 e 4 t e 2 t + 2 e 4 t # MA341 Homework 13 Solutions 33. x ( t ) = 1 2 2 2 1 2 2 2 1 x ( t ), x (0) =  2 3 2 . The characteristic equation for this matrix is A rI = 1 r 2 2 2 1 r 2 2 2 1 r = ( r + 1) 2 ( r 5) = 0 . Hence, the eigenvalues of this matrix are r 1 = r 2 = 1, and r 3 = 5. To find the eigenvectors that correspond to r = 1, we must solve ( A rI ) u = 0: 2 2 2 2 2 2 2 2 2 u 1 u 2 u 3 = . The foregoing system is equivalent to one scalar equation: u 1 u 2 + u 3 = 0 . Therefore, the solution is obtained by assigning arbitrary values to u 2 and u 3 (say, u 2 = s , u 3 = v ). Then, u 1 = s v . Consequently, the eigenvectors associated with r = 1 can be expressed as u = s v s v = s 1 1 + v  1 1 . First taking s = 1, v = 0 and then taking s = 0, v = 1, we get two linearly independent eigenvectors u 1 = 1 1 , u 2 =  1 1 . For r 3 = 5, the equation ( A r 3 I ) u = 0 reads:  4 2 2 2 4 2 2 2 4 u 1 u 2 u 3 = . Its solution is u 1 = s , u 2 = s , and u 3 = s , where s is arbitrary. Therefore, the eigenvectors associated with r 3 = 5 are u 3 = s 1 1 1 . MA341 Homework 13 Solutions Taking s = 1, we obtain u 3 = 1 1 1 ....
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This homework help was uploaded on 04/16/2008 for the course MA 341 taught by Professor Schecter during the Spring '08 term at N.C. State.
 Spring '08
 Schecter
 Differential Equations, Equations

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