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301_F10_hw8_soln

# 301_F10_hw8_soln - x = y y =-5x 2y 4 3 2 1 y...

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AME 301 Homework #8 Solutions | Fall 2010 1. Section 4.1 #14 Convert the following ODE to a rst-order system: y 00 + 15 y 0 + 50 y = 0 : Set x 1 ( t ) = y ( t ) and x 2 ( t ) = y 0 ( t ). We then have x 0 1 = x 2 ; x 0 2 = 50 x 1 15 x 2 ; or x 0 = Ax where A = 0 1 50 15 : Find eigenvalues and eigenvectors: det( A I ) = ( 15 ) + 50 = ( + 10)( + 5) = 0 : 1 = 5 ; 2 = 10 : s 1 = 1 = 5 1 ; s 2 = 1 = 10 1 : The general solution is x ( t ) = c 1 exp( 5 t ) 1 = 5 1 + c 2 exp( 10 t ) 1 = 10 1 : Note that the two eigenvectors are linearly independent, but nearly parallel (far from orthogonal). Apply the initial condition x 1 (0) = 1, x 2 (0) = 2. From the general solution x (0) = c 1 s 1 + c 2 s 2 or 1 = 5 1 = 10 1 1 c 1 c 2 = 1 2 ) c 1 c 2 = 12 14 : Note that the coe cients c 1 and c 2 have magnitudes that are very large compared to the magnitude of the initial condition x (0). This occurs because the eigenvectors are nearly parallel, so that S is nearly singular (det( S ) = 0 : 1). 2. Section 4.3 #8 Find the general solution to the system x 0 = Ax , and nd the solution satisfying the initial condition x (0) = x 0 where A = 8 1 1 10 and x 0 = 2 1 : Find eigenvalues and eigenvectors: det( A I ) = (8 )(10 ) + 1 = ( 9) 2 = 0 : 1

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1 = 2 = 9 : ( A 1 I ) x = 1 1 1 1 x 1 x 2 = 0 ) s 1 = 1 1 : The eigenvalue 1 = 9 has an algebraic multiplicity of 2, but a geometric multiplicity of 1, so the matrix A is defective. The usual analysis has produced only one linearly independent solution to the di er- ential equation, x ( t ) = c 1 exp( 1 t ) s 1 : To obtain a second linearly independent solution, we choose the form x ( t ) = ( t s 1 + u ) exp( 1 t ) ; where u is a ‘generalized’ eigenvector that is determined by the following analysis. Then x 0 ( t ) = s 1 exp( 1 t ) + ( t s 1 + u ) 1 exp( 1 t ) ; and substituting the expressions for x ( t ) and x 0 ( t ) into the di erential equation ( x 0 Ax = 0 ), we obtain s 1 exp( 1 t ) + ( t s 1 + u ) 1 exp( 1 t ) A ( t s 1 + u ) exp( 1 t ) = 0 : The exponential exp( 1 t ) is common to all terms and can be divided out. Also, note that As 1 = 1 s 1 . The remaining terms can be rearranged to form an inhomogeneous equation for u , ( A 1 I ) u = s 1 or 1 1 1 1 u 1 u 2 = 1 1 ) u = u 1 u 2 = 1 0 + 1 1 ; where is an arbitrary constant. Note that the vector multiplying in the expression for u is the eigenvector s 1 . This is expected, since s 1 lies in the null space of ( A 1 I ).
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