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Differential Equations Lecture Work Solutions 198

# Differential Equations Lecture Work Solutions 198 - A = 1 B...

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4. a. 2 u ∂x 2 +3 2 u ∂x∂y +2 2 u ∂y 2 =0 A =1 ,B =3 ,C =2 . The discriminant ∆ = 3 2 4 · 1 · 2=9 8=1 > 0 Therefore the problem is hyperbolic. The characteristic equation is dy dx = B ± 2 A = 3 ± 1 2 The Frst equation is dy dx =2 and the second is dy dx =1 Integrating, we get y =2 x + C and y = x + D Both characteristic families are straight lines. 4. b. 2 u ∂x 2 2 2 u ∂x∂y + 2 u ∂y 2 =0
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Unformatted text preview: A = 1 , B = β 2 , C = 1 . The discriminant β = ( β 2) 2 β 4 Β· 1 Β· 1 = 4 β 4 = 0 Therefore the problem is parabolic. The characteristic equation is dy dx = B Β± β β 2 A = β 2 2 = β 1 Integrating, we get y = β x + C The characteristic family is straight lines. 198...
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