Differential Equations Lecture Work Solutions 186

# Differential Equations Lecture Work Solutions 186 - Now we...

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Now we can compute the new coeﬃcients or compute each of the derivative in the equa- tion. We chose the latter. u xx = u ξξ ( 1 1+4 x ) 2 +2 u ξη ( 1 1+4 x )( 1+ 1+4 x ) + u ηη ( 1+ 1+4 x ) 2 + u ξ [ 2(1 + 4 x ) 1 / 2 ]+ u η [2 (1 + 4 x ) 1 / 2 ] = u ξξ [1 + 2 1+4 x +1+4 x ]+2 u ξη (1 (1 + 4 x )) + u ηη [1 2 1+4 x +1+4 x ] 2(1 + 4 x ) 1 / 2 u ξ +2 (1+4 x ) 1 / 2 u η u xy =2 u ξξ ( 1 1+4 x )+ u ξη [2 ( 1 1+4 x )+2( 1+ 1+4 x )] + u ηη 2( 1+ 1+4 x ) u yy =4 u ξξ +2 u ξη · 4+ u ηη · 4 u xx + u xy xu yy = u ξξ [2 + 4 x +2 1+4 x ]+2 u ξη ( 4 x )+ u ηη (2 + 4 x 2 1+4 x ) 2(1 + 4 x ) 1 / 2 u ξ +2(1+4 x ) 1 / 2 u η +2 u ξξ ( 1 1+4 x )+ u ξη ( 4) + 2 u ηη ( 1+ 1+4 x ) 4 x ( u ξξ +2 u ξη + u ηη )= (2 + 4 x +2 1+4 x 2 2 1+4 x 4 x ) u
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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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