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Unformatted text preview: ( t ) = − 4 c 1 e − 4 t + 2 c 2 e 2 t and solve the system y (0) = c 1 e − 4 · + c 2 e 2 · = c 1 + c 2 = 3 , y (0) = − 4 c 1 e − 4 · + 2 c 2 e 2 · = − 4 c 1 + 2 c 2 = − 12 ⇒ c 1 = 3 , c 2 = 0 . Therefore, the solution to the given initial value problem is y ( t ) = (3) e − 4 t + (0) e 2 t = 3 e − 4 t . 15. The auxiliary equation for this equation is r 2 + 2 r + 1 = ( r + 1) 2 = 0. We see that r = − 1 is a repeated root. Thus, two linearly independent solutions are y 1 ( t ) = e − t and y 2 ( t ) = te − t . This means that a general solution is given by y ( t ) = c 1 e − t + c 2 te − t . 170...
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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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