351sechom_a - Solutions for Second Order Examples(1 d 2 y...

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Unformatted text preview: Solutions for Second Order Examples (1) d 2 y dx 2- dy dx = 2 y . Characteristic equation: λ 2- λ- 2 = 0 Characteristic roots: λ 1 =- 1 , λ 2 = 2 General Solution: y ( x ) = c 1 e- t + c 2 e 2 t (2) y 00- 16 y = 0 , with the two sets of initial conditions ( a ) y (1) = 1 y (1) = 0 , and ( b ) y (1) = 0 y (1) = 1 . For (a): Characteristic equation: λ 2 + 16 = 0 Characteristic roots: λ 1 =- 4 , λ 2 = 4 General Solution: y ( x ) = c 1 e- 4 x + c 2 e 4 x First Derivative: y ( x ) =- 4 c 1 e- 4 x + 4 c 2 e 4 x Apply Initial Conditions: y (1) = e- 4 c 1 + e 4 c 2 y (1) =- 4 e- 4 c 1 + 4 e 4 c 2 Equations for Coefficients: e- 4 c 1 + e 4 c 2 = 1- 4 e- 4 c 1 + 4 e 4 c 2 = 0 Coefficients: c 1 = 1 2 e 4 , c 2 = 1 2 e- 4 Solution of IVP: y ( x ) = 1 2 e 4 e- 4 x + 1 2 e- 4 e 4 x or y ( x ) = cosh (4 x- 4) . 1 For (b): Characteristic equation: λ 2 + 16 = 0 Characteristic roots: λ 1 =- 4 , λ 2 = 4 General Solution: y ( x ) = c 1 e- 4 x + c 2 e 4 x First Derivative: y ( x ) =- 4 c 1 e- 4 x...
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351sechom_a - Solutions for Second Order Examples(1 d 2 y...

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