chapter3.page07

chapter3.page07 - t = 0 we have x 1 (0) = C 1 e + C 2 e...

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1. VECTORS AND MATRICES 7 Notice, by default, Mathcad will display the results as dec- imal, you can double click on the result vector to change it to fraction, after you double click the result you will have a popup menu such as Figure 1. Format Result a The next example shows how we can determine the unknown con- stants typically found in the initial value problems of system of differ- ential equations. Example 1.5 . Let x 1 ( t ) = C 1 e t + C 2 e 2 t and x 2 = 2 C 1 e t - C 2 e 2 t . If x 1 (0) = 2 , x 2 (0) = 3 , find C 1 and C 2 Solution From x 1 ( t ) = C 1 e t + C 2 e 2 t , set
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Unformatted text preview: t = 0 we have x 1 (0) = C 1 e + C 2 e 2(0) = C 1 + C 2 . Similarly, x 2 ( t ) = 2 C 1 e t-C 2 e 2 t , gives x 2 (0) = 2 C 1 e ( 0)-C 2 e 2(0) = 2 C 1-C 2 . Together with x 1 (0) = 2 , x 2 (0) = 3 we have the following system of equations, C 1 + C 2 = 1 2 C 1-C 2 = 3 Rewrite the equation in matrix form 1 1 2-1 C 1 C 2 = 1 3 , and using Mathcad we nd the solution is C 1 C 2 = 4 3-1 3...
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This note was uploaded on 10/04/2011 for the course MAP 4231 taught by Professor Dr.han during the Spring '11 term at UNF.

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