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Unformatted text preview: x (1) ( t ) = " 1 1 # et , x (2) ( t ) = " 41 # e6 t , and the general solution is x ( t ) = c 1 x (1) ( t ) + c 2 x (2) ( t ) = c 1 " 1 1 # et + c 2 " 41 # e6 t . To obtain the solution satisfying the initial conditions, c 1 and c 2 must satisfy x (0) = " 32 # = c 1 " 1 1 # + c 2 " 41 # or 3 = c 1 + 4 c 2 ,2 = c 1c 2 . Solving these gives c 1 =1, c 2 = 1, and the desired solution is x ( t ) = (1) " 1 1 # et + (1) " 41 # e6 t = "11 # et + " 41 # e6 t . 1. Find the general solution of d x dt = " 2 34 # x . 2. Find the general solution of d x dt = " 2 625 # x ....
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This note was uploaded on 12/02/2009 for the course MATH 352 taught by Professor Staff during the Spring '08 term at University of Delaware.
 Spring '08
 Staff
 Eigenvectors, Vectors

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