Unformatted text preview: x p t ( ) = C , must resemble form of righthand side, which is 0 plug into equation to find C : C + 5 C + 6 C = 6 , C = 1 , x p t ( ) = 1 Genera solution is sum of two. x t ( ) = x h t ( ) + x p t ( ) = A 1 e − 2 t + A 2 e − 3 t + 1 Now plug in initial conditions x ( ) = A 1 e − 2 ⋅ + A 2 e − 3 ⋅ + 1 = A 1 + A 2 + 1 = x ( ) = − 2 A 1 e − 2 t t = − 3 A 2 e − 3 t t = = − 2 A 1 + − 3 A 2 = A 1 = − 3 2 A 2 , − 3 2 A 2 + A 2 + 1 = − 1 2 A 2 = − 1 , A 2 = 2 , A 1 = − 3 a. x t ( ) = − 3 e − 2 t + 2 e − 3 t + 1 for t ≥ b. For an n th order differential equation, the characteristic equation will have n roots, and so there will be n (complex) exponentials in the homogeneous solution. c. (iv) After. Always remember to plug in initial conditions at the very end....
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 Fall '11
 KUO
 Exponential Function, Quadratic equation, Complex number, initial conditions

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