Math 3301 Homework Set Points ( ) ( ) I ll leave it to you to verify that the eigenvalues and eigenvectors for this matrix are, ( ) ( ) ( ) ( )

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1 #7. ( pts) I ll leave it to you to verify that the eigenvalues and eigenvectors for this matrix are, λ 5 λ 7 t t ce The general solution is then : 5 7 c c c x( 0) c c 9 9 c+ c c t 5t 7 e + e A sketch of the phase portrait is to the right. The equilibrium solution in this case is an unstable node. #0. ( pts) Here s the system x y x x 0 x x x( 0) x y x 7x+ x 7 8 I ll leave it to you to verify that the eigenvalues and eigenvectors for this matrix are, λ i + λ i + i i ( + it ) t e e ( cos( t) + isin ( t) ) + i + i t cos t sin e ie cos t sin t cos t + sin t t cos t sin x ce cos t sin t cos t + sin t 0 c x( 0) c c 8 c 5 8 c+ c c

2 5 t cos 5 t sin e e cos t sin t cos t + sin t t cos sin e 8cos sin The solution to the original IVP is then, t 5 t e cos e sin ( ) y t t t axis going downwards. A sketch of the phase portrait is to the right. The equilibrium solution in this case is an unstable spiral. #. ( pts) I ll leave it to you to verify that the eigenvalues and eigenvectors for this matrix are, λ, ρ ρ ρ ρ ρ ρ + + ρ 0 The general solution is then : t t t ce c te e c+ c x( 0) c c 0 c c c t 8 t t t t e te e e 0 + 8t

3 6 6 0 axis going upwards. A sketch of the phase portrait is to the right. The equilibrium solution in this case is an asymptotically stable improper node. #. First the eigenvalues. Not Graded λ det ( A λi) ( λ)( λ) + 5 λ + λ+ 8 λ, ± i 5 λ Now the eigenvector for λ + i i 0 ( i) 0 ( i) 5 i i Then by the fact given in class the eigenvector for λ i #. First the eigenvalues. 6 i λ det ( A λi) ( λ)( λ) + ( λ ) λ λ The (only) eigenvector for λ, is then, is, 6, #. First the eigenvalues. λ det ( A λi) ( λ)( λ) ( λ+ )( λ 6) λ, λ 6 7 λ Now the eigenvector for λ the eigenvector for λ 6.

4 #. #5. x y x x 0 x x x( 0) 5 5 x y x 5x+ x 5 x y x x 0 0 x y x x x 0 0 x x( 0) 6 x y x x 9x #6. I ll leave it to you to verify that the eigenvalues and eigenvectors for this matrix are, λ λ ( 6+ 5) t ( 6 5) t x ce #8. I ll leave it to you to verify that the eigenvalues and eigenvectors for this matrix are, 7 λ 9 λ 9t t 7 x ce 7 c+ 7c c x( 0) c c 0 0 c+ c c

5 t 7 9t e + e A sketch of the phase portrait is to the right. The equilibrium solution in this case is an unstable saddle point. #9. I ll leave it to you to verify that the eigenvalues and eigenvectors for this matrix are, + i i λ i λ i it + i + i ( cos( ) sin ( )) cos t sin t cos t sin t e t + i t i cos sin The general solution is then : cos t sin t cos t sin t c c cos sin c + c c x( 0) c c 0 c c ( 7 t) ( t) + cos t sin t cos t sin t cos sin cos sin cos sin 5 0 axis going upwards. A sketch of the phase portrait is to the right. The equilibrium solution in this case is a stable center. #. I ll leave it to you to verify that the eigenvalues and eigenvectors for this matrix are,

6 λ, 6 ρ 6 9 ρ ρ 6 ρ 9+ 6ρ ρ 9 ρ 0 t 6 t 6 t 9 x ce c te e c + 9c x( 0) c c 0 c 8 c c t t t e t t t e e e t axis going upwards. A sketch of the phase portrait is to the right. The equilibrium solution in this case is an unstable improper node.

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