Differential Equations Lecture Work Solutions 181

Differential Equations Lecture Work Solutions 181 - 1a....

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1a. xu xx + u yy = x 2 A = xB =0 C =1 4 = B 2 4 AC = 4 x If x> 0then 4 < 0 elliptic = 0 = 0 parabolic < 0 > 0 hyperbolic characteristic equation dy dx = ± 4 x 2 x = ± x x Suppose x< 0 (hyperbolic) Let z = x (then z> 0) then dz = dx and dy dz = dy dx = ± z z = ± 1 z dy = ± dz z 1 / 2 y = ± 2 z + c y 2 z = c characteristic curves: y 2 z = c 2 families as expected. Transformation: ξ = y 2 z η = y +2
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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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