Unformatted text preview: 2 = − 4 . Thus, r = − 4 is a double root of the characteristic equation. So, a general solution has the form y = ( C 1 t + C ) e − 4 t . For constants C 1 and C 2 , we obtain the system y (0) = ( C 1 t + C ) e − 4 t ± ± t =0 = C = 1 , y (0) = ( − 4 C 1 t − 4 C + C 1 ) e − 4 t ± ± t =0 = C 1 − 4 C = 0 ⇒ C = 1 , C 1 = 4 , and so y ( t ) = (4 t + 1) e − 4 t . 234...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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