Differential Equations Lecture Work Solutions 274

Differential Equations Lecture Work Solutions 274 - 3 2.5 2...

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Unformatted text preview: 3 2.5 2 1.5 1 u=2 0.5 u=0 u=1 0 −3 −2 −1 0 1 2 3 4 Figure 54: Solution for 7 [u] = 2 − 0 = 2 Thus dxs =1 dt and the characteristic is then xs = t + C. To find C , we substitute the point of intersection t = 1 and xs = 3 . Thus 2 3 =1+C 2 or C= 1 2 The third shock characteristic is then xs = t + 1 . 2 The shock characteristics and the solutions in each domain are given in the figure above. 274 ...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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