Differential Equations Lecture Work Solutions 182

# Differential Equations Lecture Work Solutions 182 - z = 2 1...

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η z =2 1 2 z 1 / 2 ± = 1 z η x = 1 z ξ y =1 η y =1 ξ xx =( ξ x ) x = 1 z ! x = 1 z ! z z x = 1 2 z 3 / 2 ± = 1 2 z 3 / 2 η xx =( η x ) x = 1 z ! x = 1 z ! z z x = 1 2 z 3 / 2 ± = 1 2 z 3 / 2 ξ xy = ξ yy = η xy = η yy =0 u xx = 1 z u ξξ 2 z u ξη + 1 z u ηη + 1 2 z 3 / 2 u ξ 1 2 z 3 / 2 u η u yy = u ξξ ξ 2 y |{z} =1 +2 u ξη ξ y η y + u ηη η 2 y |{z} =1 + u ξ ξ yy |{z} =0 + u η η yy |{z} =0 = u ξξ +2 u ξη + u ηη Substitute in the equation x |{z} z ² 1 z u ξξ 2 z u ξη + 1 z u ηη + 1 2 z 3 / 2 u ξ 1 2 z 3 / 2 u η ³ + u ξξ +2 u ξη + u ηη = x 2 |{z} ( z ) 2 u ξξ +2 u ξη u ηη 1 2 z 1 / 2 u ξ + 1 2 z 1 / 2 u η + u ξξ +2 u ξη u ηη = z 2 4 u ξη 1
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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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