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Unformatted text preview: y ( t ) = [ ( 1 + 5 ) / 2] e ( 5+ 5) t + [ ( 1 5 ) / 2] e ( 5 5) t is the solution to the initial value problem. k = 25 k = 25 = 25. Then r = 5 25 25 = 5. Thus, r = 5 is a double root of the characteristic equation. So, a general solution has the form y = ( C 1 t + C ) e 5 t . For constants C 1 and C 2 , using the initial conditions, we obtain the system y (0) = ( C 1 t + C ) e 5 t t =0 = C = 1 , y (0) = ( 5 C 1 t 5 C + C 1 ) e 5 t t =0 = C 1 5 C = 0 C = 1 , C 1 = 5 , and so y ( t ) = (5 t + 1) e 5 t . 235...
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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 University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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