228_pdfsam_math 54 differential equation solutions odd

228_pdfsam_math 54 differential equation solutions odd - a...

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Chapter 4 Substituting these values into equation (4.8) we obtain y (2) e 1 2 e 2 1 e 2 2 (3 . 0592) + e 2 2 (0 . 1706) = 1 . 9271 . Repeating these calculations for n = 3, 4, and 5 yields the approximations in Table 4-A. Table 4–A : Successive approximations for y (2) using Simpson’s rule. Intervals y (2) y (2) y 6 1 . 9275 8 1 . 9275 10 1 . 9275 Since these values do not change in the third place, we can expect that the Frst three places are accurate and we obtained an approximate solution of y (2) = 1 . 93 . 21. A particular solution to the given equation has the form y p ( t )= v 1 ( t ) y 1 ( t )+ v 2 ( t ) y 2 ( t )= v 1 ( t ) e t + v 2 ( t )( t +1) . Since y 0 1 ( t )= e t , y 0 2 ( t ) 1, the system (9), with
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Unformatted text preview: a = a ( t ) = t and g ( t ) = t 2 , becomes v 1 ( t ) e t + v 2 ( t )( t + 1) = 0 , v 1 ( t ) e t + v 2 ( t ) = t 2 t = t. Subtracting the second equation from the Frst one, we get tv 2 ( t ) = t v 2 ( t ) = 1 v 2 ( t ) = t. Substituting v 2 ( t ) into the Frst equation yields v 1 ( t ) e t ( t + 1) = 0 v 1 ( t ) = ( t + 1) e t v 1 ( t ) = Z ( t + 1) e t dt = ( t + 1) e t + Z e t dt = ( t + 2) e t . 224...
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