226_pdfsam_math 54 differential equation solutions odd

226_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 Adding a general solution to the homogeneous equation, we get y ( t )= c 1 cos 2 t + c 2 sin 2 t 1 2 cos 2 t ln | sec 2 t +tan2 t |− 1 5 e t . 19. A general solution of the corresponding homogeneous equation is given by y h ( t )= c 1 e t + c 2 e t . We will try to Fnd a particular solution to the original nonhomogeneous equation of the form y p ( t )= v 1 ( t ) y 1 ( t )+ v 2 ( t ) y 2 ( t ), where y 1 ( t )= e t and y 2 ( t )= e t . We apply formulas (10) on page 195 in the text, but replace indeFnite integrals by deFnite integrals. Note that y 1 ( t ) y 0 2 ( t ) y 0 1 ( t ) y 2 ( t )= e x e x ( e x ) e x =2 . With g ( t )=1 /t and integration from 1 to t , formulas (10) yield v 1 ( t )= t Z 1 g ( x ) y 2 ( x ) 2 dx = 1 2 t Z 1 e x x dx , v 2 ( t )= t Z 1 g ( x ) y 1 ( x ) 2 dx = 1 2 t Z 1 e x x dx . (Notice that we have chosen the lower limit of integration to be equal to 1 because the initial conditions are given at 1. We could have chosen any other value for the lower limit, but the
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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...
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