Differential Equations Lecture Work Solutions 184

Differential Equations Lecture Work Solutions 184 - Again,...

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Again, substitute for x: 2 x = β x = 1 2 β x = 1 4 β 2 u αα + u ββ + 1 2 1 1 2 β u β = 1 4 β 2 ± 2 u αα + u ββ = 1 β u β + 1 16 β 4 For the parabolic case x =0 the equation becomes: 0 · u xx + u yy =0 or u yy =0 which is already in a canonical form This parabolic case can be solved.
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