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hw6-solutions

# hw6-solutions - Mathematics 53 Winter 2011 Solutions to...

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Mathematics 53, Winter 2011 Solutions to Homework 6 Section 3.4. Problem 8: We consider the diferential equation vy ( t ) = b 2 - 5 1 - 2 B ( t ) with initial condition (0) = b 3 2 B . We begin by computing the eigenvalues oF the coe±cient matrix A : 0 = det b λ - 2 5 - 1 λ + 2 B = ( λ - 2)( λ + 2) + 5 = λ 2 + 1 . ThereFore, the eigenvalues are i and - i . The vector vV = b 2 + i 1 B is an eigenvector oF A with eigenvalue i . Consequently, the Function vx ( t ) = e it b 2 + i 1 B is a solution oF the diferential equation ( t ) = Avx ( t ). We now split ( t ) into its real and imaginary parts: ( t ) = (cos( t ) + i sin( t )) b 2 + i 1 B = b 2cos( t ) - sin( t ) cos( t ) B + i b cos( t ) + 2sin( t ) sin( t ) B . ThereFore, the Functions vu ( t ) = b t ) - sin( t ) cos( t ) B , vw ( t ) = b cos( t ) + 2sin( t ) sin( t ) B Form a Fundamental system oF solutions. In other words, the general solution oF the diferential equation v y ( t ) = Avy ( t ) is given by ( t ) = a 1 b t ) - sin( t ) cos( t ) B + a 2 b cos( t ) + 2sin( t ) sin( t ) B

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for suitable constants a 1 , a 2 . The initial condition v y (0) = b 3 2 B implies a 1 = 2 and a 2 = - 1. Thus, vy ( t ) = b 3cos( t ) - 4sin( t ) 2cos( t ) - sin( t ) B . Problem 9: We consider the diFerential equation ( t ) = b 1 - 5 1 - 3 B ( t ) with initial condition (0) = b 1 1 B . We begin by computing the eigenvalues of the coe±cient matrix A : 0 = det b λ - 1 5 - 1 λ + 3 B = ( λ - 1)( λ + 3) + 5 = λ 2 + 2 λ + 2 . Therefore, the eigenvalues are - 1 + i and - 1 - i . The vector vV = b 2 + i 1 B is an eigenvector of A with eigenvalue i . Consequently, the function vx ( t ) = e ( 1+ i ) t b 2 + i 1 B is a solution of the diFerential equation ( t ) = Avx ( t ). We now split ( t ) into its real and imaginary parts: ( t ) = e t (cos( t ) + i sin( t )) b 2 + i 1 B = e t b t ) - sin( t ) cos( t ) B + i e t b cos( t ) + 2sin( t ) sin( t ) B . Therefore, the functions vu ( t ) = e t b t ) - sin( t ) cos( t ) B , vw ( t ) = e t b cos( t ) + 2sin( t ) sin( t ) B
form a fundamental system of solutions. In other words, the general solution of the diFerential equation v y ( t ) = Avy ( t ) is given by v y ( t ) = a 1 e t b 2cos( t ) - sin( t ) cos( t ) B + a 2 e

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hw6-solutions - Mathematics 53 Winter 2011 Solutions to...

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