Differential Equations Lecture Work Solutions 185

Differential Equations Lecture Work Solutions 185 - 1b. uxx...

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1b. u xx + u xy xu yy =0 A =1 B C = x 4 =1+4 x> 0i f 1 4 hyperbolic = 1 4 parabolic < 0 < 1 4 elliptic dy dx = 1 ± 1+4 x 2 Consider the hyperbolic case: 2 dy =(1 ± x ) dx Integrate to get characteristics 2 y = x ± 2 3 · 1 4 (1 + 4 x ) 3 / 2 + c 2 y x 1 6 (1 + 4 x ) 3 / 2 = c ξ =2 y x 1 6 (1 + 4 x ) 3 / 2 η y x + 1 6 (1 + 4 x ) 3 / 2 ξ x = 1 1 6 · 3 2 · 4(1 + 4 x ) 1 / 2 = 1 x ξ xx = 1 2 (1 + 4 x ) 1 / 2 · 4= 2(1 + 4
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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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