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Differential Equations Lecture Work Solutions 57

Differential Equations Lecture Work Solutions 57 - Take ln...

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Take ln of both sides 2 π λ = 0 2 π λ = 0 λ = 0 λ = 0 not possible not possible Thus trivial solution if λ < 0 4. 1 r ∂ r r ∂u ∂ r + 1 r 2 2 u ∂ θ 2 = 0 u ( a, θ ) = f ( θ ) u ( r, 0) = u θ ( r, α ) = 0 u ( r, θ ) = R ( r ) Θ ( θ ) Θ 1 r ∂ r r ∂R ∂ r + 1 r 2 R 2 Θ ∂ θ
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