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Differential Equations Lecture Work Solutions 249

Differential Equations Lecture Work Solutions 249 - 2 u((0...

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Unformatted text preview: 2 u(η (0), 0) = 1 + cos α. 5 The solution of the first ODE, gives the characteristics in the transformed domain: η= 1 ξ+α 2 The solution of the second ODE: 1 ξ u(η (ξ ), ξ ) = Ke 2 Using the initial condition 2 1 + cos α = K 5 Thus 1 But α = η − ξ thus 2 Now substitute back: Thus 1 ξ 2 u(η (ξ ), ξ ) = (1 + cos α)e 2 5 1 ξ 1 2 u(η (ξ ), ξ ) = (1 + cos (η − ξ ))e 2 5 2 1 1 ξ = x+t 2 2 1 5 1 η − ξ = (t − 2x) − ( x + t) = x 2 2 2 1 x+t u(x, t) = (1 + cos x)e 2 . 249 ...
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