Differential Equations Lecture Work Solutions 188

Differential Equations Lecture Work Solutions 188 - u ξ ξ...

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1c. x 2 u xx +2 xy u xy + y 2 u yy + xy u x + y 2 u y =0 A = x 2 B =2 xy C = y 2 4 =4 x 2 y 2 4 x 2 y 2 = 0 parabolic dy dx = 2 xy 2 x 2 = y x dy y = dx x ξ = `n y `n x ξ = `n y x ± e ξ = y x η = x arbitrarily chosen since this is parabolic ξ x = 1 x ξ y = 1 y ξ xx = 1 x 2 ξ xy =0 ξ yy = 1 y 2 η x =1 η y = η xx = η xy = η yy =0 u xx = 1 x 2 u ξξ +2 u ξη ( 1 x )+ u ηη + 1 x 2 u ξ u xy = 1 xy u ξξ + u ξη 1 y u yy = 1 y 2 u ξξ 1 y 2 u ξ u ξξ 2 xu ξη + x 2 u ηη + u ξ 2 u ξξ +2 xu ξη
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Unformatted text preview: + u ξ ξ − u ξ + xy ( − 1 x u ξ + u η ) + y 2 ( 1 y u ξ ) = 0 x 2 u η η + xyu η = 0 u η η = − e ξ u η y = e ξ x therefore y/x = e ξ This equation can be solved. 188...
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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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