# Thus u t 5 e 5 t v 1 e 2 t v 2 solves the given

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Thus u ( t ) = 5 e 5 t v 1 e 2 t v 2 solves the given diferential equation. Conse- quently, u (2) = b 5 e 10 + 4 e 4 5 e 10 8 e 4 B . 022 10.0 points ±ind the solution oF the diferential equa- tion d u dt = A u ( t ) , u (0) = b 12 16 B
ramirez (br23624) – HW 10 – gilbert – (53415) 15 when A is the matrix A = 1 3 b 17 8 16 7 B 1. u ( t ) = b 8 e 3 t + 4 e 1 3 t 8 e 3 t + 8 e 1 3 t B 2. u ( t ) = b 8 e 3 t 4 e 1 3 t 8 e 3 t 8 e 1 3 t B correct 3. u ( t ) = b 8 e 3 t 4 e 1 3 t 8 e 3 t 8 e 1 3 t B 4. u ( t ) = b 8 e 3 t + 4 e 1 3 t 8 e 3 t + 8 e 1 3 t B 5. u ( t ) = b 8 e 3 t + 4 e 1 3 t 8 e 3 t 8 e 1 3 t B 6. u ( t ) = b 8 e 3 t 4 e 1 3 t 8 e 3 t + 8 e 1 3 t B Explanation: Since det[ A λI ] = v v v v v 17 3 λ 8 3 16 3 7 3 λ v v v v v = ( 17 3 λ )( 7 3 λ ) + 128 = λ 2 10 3 λ + 1 = ( λ 3)( λ 1 3 ) , the eigenvalues of A are λ 1 = 3, λ 2 = 1 3 and corresponding eigenvectors v 1 = b 2 2 B , v 2 = b 1 2 B form a basis for R 2 because λ 1 n = λ 2 . Thus u (0) = c 1 v 1 + c 2 v 2 . To compute c 1 and c 2 we apply row reduction to the augmented matrix [ v 1 v 2 u (0)] = b 2 1 12 2 2 16 B b 1 0 4 0 1 4 B . This shows that c 1 = 4, c 2 = 4 and u (0) = 4 v 1 4 v 2 . Since v 1 and v 2 are eigenvectors correspond- ing to the eigenvalues 3 and 1 3 respectively, set u ( t ) = 4 e 3 t v 1 4 e 1 3 t v 2 . Then u (0) is the given initial value and A u ( t ) = 4 e 3 t A v 1 4 e 1 3 t A v 2 = 4 ( 3 e 3 t ) v 1 4 p 1 3 e 1 3 t P v 2 = d u ( t ) dt . This shows that u ( t ) solves the diFerential equation. But u ( t ) = 4 e 3 t b 2 2 B 4 e 1 3 t b 1 2 B = b 8 e 3 t 4 e 1 3 t 8 e 3 t 8 e 1 3 t B . Consequently, u ( t ) = b 8 e 3 t 4 e 1 3 t 8 e 3 t 8 e 1 3 t B solves the given diFerential equation.