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Unformatted text preview: et . (5 points). 1. The characteristic equation is 0 = r 2 + 2 r3 = ( r + 3)( r1). Thus the general solution is y = C 1 e3 t + C 2 e t . 2. This would correspond to a characteristic equation of 0 = ( r5)( r + 1) = r 24 r5, so a corresponding dierential equation would be y 004 y5 y = 0....
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 Fall '07
 COSTIN
 Math, Differential Equations, Equations

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