This preview shows page 1. Sign up to view the full content.
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 dierential 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...
View Full
Document
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details