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Unformatted text preview: ( x ) = exp Z 1 x dx = e ln x = 1 x , for x > . Multiplying the equation by this integrating factor yields 1 x dy dx y x 2 = e x D x y x = e x . Integrating gives y x = e x + C y = xe x + Cx. Now applying the initial condition, y (1) = e 1, we have e 1 = e + C C = 1 . 44...
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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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