Unformatted text preview: . Therefore the general solution to the system of diﬀerential equations is given by y = c 1 ± 11 ² e2 t + c 2 ± 1 1 ² e 6 t . From the initial condition, we have that ± 1 ² = c 1 ± 11 ² + c 2 ± 1 1 ² , and therefore c 1 =1 2 and c 2 = 1 2 . Therefore the solution to the initial value problem is given by y =1 2 ± 11 ² e2 t + 1 2 ± 1 1 ² e 6 t . (b) What type of system is the above diﬀerential equation? Circle one. i. source ii. sink iii. saddle point iv. spiral in v. spiral out vi. periodic Since the eigenvalues are real and have opposite signs, it is a saddle point. 1...
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 Spring '10
 11
 Linear Algebra, Complex number, GSI Carter

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