Unformatted text preview: = c 1 e r (0) + c 2 e s (0) = c 1 + c 2 y ± ( t ) = rc 1 e rt + sc 2 e st y ± (0) = rc 1 e r (0) + sc 2 e s (0) = rc 1 + sc 2 Hence c 1 = y ± (0) − sy (0) r − s , c 2 = y ± (0) − ry (0) s − r and the particular solution is y ( t ) = ± y ± (0) − sy (0) r − s ² e rt + ± y ± (0) − ry (0) s − r ² e st which satisFes the initial conditions y (0) and y ± (0). Example 2.17 Suppose d 2 y dt 2 + 4 dy dt − 5 y = Then the auxiliary equation is x 2 + 4 x − 5 = ( x + 5)( x − 1) = Hence, r = − 5 and s = 1, with the general solution y ( t ) = c 1 e − 5 t + c 2 e t...
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 Fall '11
 Dr.Gwartney
 Economics, Quadratic equation, homogeneous equation, auxiliary equation

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