Unformatted text preview: choice of 1 will make the determination of the constants c 1 and c 2 easier.) Thus y p ( t ) = e t 2 t Z 1 e − x x dx − e − t 2 t Z 1 e x x dx , and so a general solution to the original di±erential equation is y ( t ) = c 1 e − t + c 2 e t + e t 2 t Z 1 e − x x dx − e − t 2 t Z 1 e x x dx . By plugging in the Frst initial condition (and using the fact that the integral of a function from a to a is zero which is why we have chosen the lower limit of integration to be the initial point, t = 1), we Fnd that y (1) = c 1 e − 1 + c 2 e 1 = 0 . 222...
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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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