Differential Equations Lecture Work Solutions 224

Differential Equations Lecture Work Solutions 224 - 1c. uxx...

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1c. u xx + u xy =0 A =1 B C 4 = 1 hyperbolic dy dx = +1 ± 1 2 % +1 & 0 y =+ x + K 1 y = K 2 ξ = y x = 1 ξ y η = x η y u x = u ξ + u η η x |{z} =0 = u ξ u y = u ξ + u η u xx = u ξξ u xy = u u ξη u xx + u xy = u The solution in the original domain is then:
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