hw5key - AMATH 351 Homework 5 Section7.1 6 Section7,2 15 21...

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Unformatted text preview: AMATH 351 Homework 5 Section7.1 6 Section7,2 15, 21, 22 Section7.3 22 Section7.5 2, 11, 31 Section 7.1 6 u 00 + p ( t ) u + q ( t ) u = g ( t ) , u (0) = u , u (0) = u Let u 1 = u, u 2 = u , then the di erential eqn can be written as u 2 + p ( t ) u 2 + q ( t ) u 1 = g ( t ) So the initial value problem is equivalent to ( u 1 = u 2 u 2 =- q ( t ) u 1- p ( t ) u 2 + g ( t ) u 1 (0) = u , u 2 (0) = u Section 7.2 15 1 2 1 1 2 1 1 2 1 → 1 1 2 1 2 1 1 2 1 2 1 1 2 → 1- 1 4 1 2- 1 4 1 1 2 1 2 1 1 2 → 1 1 2- 1 4 1 8 1 1 2 1 2 1 1 2 → 1 1 2- 1 4 1 8 1 1 2- 1 6 1 1 2 So the inverse matrix is 1 2- 1 4 1 8 1 2- 1 6 1 2 21 A ( t ) = e t 2 e- t e 2 t 2 e t e- t- e 2 t- e t 3 e- t 2 e 2 t , B ( t ) = 2 e t e- t 3 e 2 t- e t 2 e- t e 2 t 3 e t- e- t- e 2 t (a) A + 3 B = e t 2 e- t e 2 t 2 e t e- t- e 2 t- e t 3 e- t 2 e 2 t + 3 × 2 e t e- t 3 e 2 t...
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hw5key - AMATH 351 Homework 5 Section7.1 6 Section7,2 15 21...

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