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Differential Equations Lecture Work Solutions 274

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

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-3 -2 -1 0 1 2 3 4 0 0.5 1 1.5 2 2.5 3 u=0 u=1 u=2 Figure 54: Solution for 7 [ u ] = 2 0 = 2 Thus dx s dt = 1 and the characteristic is then x s = t + C. To find C , we substitute the point of intersection t = 1 and x s = 3 2 . Thus 3 2 = 1 + C or C = 1 2 The third shock characteristic is then
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