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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 Fall '08
 BELL,D
 Characteristic polynomial, straight lines, Complex number, dy, characteristic equation

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