Differential Equations Lecture Work Solutions 207

Differential Equations Lecture Work Solutions 207 - 1 3 u =...

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u ξη = 1 3 u η 8 9 This equation can be solved as follows: Let ν = u η then u ξη = ν ξ ν ξ = 1 3 ν 8 9 This is Linear 1 st order ODE ν 0 1 3 ν = 8 9 Integrating factor is e 1 3 ξ ( νe 1 3 ξ ) 0 = 8 9 e 1 3 ξ νe 1 3 ξ = 8 9 Z e 1 3 ξ d ξ = 8 3 e 1 3 ξ + C ( η ) ν = 8 3 + C ( η ) e 1 3 ξ To Fnd u we integrate with respect to η u η = 8 3 + C ( η ) e 1 3 ξ u = 8 3 η + e 1 3 ξ c 1 ( η ) | {z } integral of C ( η ) + K ( ξ ) To check the solution, we di±erentiate it and substitute in the canonical form:
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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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