Differential Equations Lecture Work Solutions 130

Differential Equations Lecture Work Solutions 130 - 3 The...

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3. The equation becomes u θθ +cot θu θ + 1 sin 2 θ u ϕϕ =0 , 0 <θ<π , 0 <ϕ< 2 π Using separation of variables u ( θ, ϕ )=Θ( θ )Φ( ϕ ) Θ 0 Φ+cot θ Θ 0 Φ+ 1 sin 2 θ ΘΦ 0 =0 Divide by ΦΘ and multiply by sin 2 θ we have sin 2 θ Θ 0 Θ +cos θ sin θ Θ 0 Θ = Φ 0 Φ = µ Thus Φ 0 + µ Φ=0 sin 2 θ Θ 0 +sin θ cos θ Θ 0 µ Θ=0 Because of periodicity, the Φ equation has solutions Φ m =
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