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# Sln1 - Name(20 1 Solve the following initial value problem...

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Name: Page 1 (20) 1. Solve the following initial value problem: z 0 - sin ( x ) z = x 2 ; z (0) = 1 The Integrating factor is : μ ( x ) = e - R sin ( x ) dx = e cos ( x ) therefore the equation can be written as d dx e cos ( x ) z = x 2 e cos ( x ) and integrating we obtain: e cos ( x ) z = Z x 2 e cos ( x ) dx + c Hence z = e - cos ( x ) Z x 2 e cos ( x ) dx + ce - cos ( x ) but z (0) = 1, thus z = e - cos ( x ) Z x 0 ξ 2 e cos ( ξ ) + e 1 - cos ( x ) or, z = Z x 0 ξ 2 e cos ( ξ ) - cos ( x ) + e 1 - cos ( x )

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Name: Page 2 (30) 2. Given the system of two first order ODEs x 1 x 2 ! 0 = 0 1 ω 2 - i 2 ω ! x 1 x 2 ! (1) (Note: i = - 1 ) (a) Can you find two linearly independent solutions by diagonalizing the system? (5pt) No because the matrix of the coefficient has one eigenvalue λ = - with multeplicity 2, and only one belonging vector q 1 = ( iω, ω 2 ) T (b) If not, How would you proceed to find the two linearly independent solutions? (5pt) We can get the second vector by solving the following system: 1 ω 2 - ! q 1 , 1 q 1 , 2 ! = ω 2 ! (2) which has solution q 1 , 1 = (1 , 0) T and reduce the matrix in Jordan Q - 1 AQ = J , where the modal matrix is Q = 1
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Sln1 - Name(20 1 Solve the following initial value problem...

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