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Unformatted text preview: Chapter 7 Using linearity of the inverse Laplace transform and formula (6) in Section 7.6, we get x ( t ) = 1 2 + 2 3 e t 1 6 e 2 t 1 4 x 2 1 3 e x + 1 12 e 2 x x = t 3 u ( t 3) = 1 2 + 2 e t 3 e 2 t 6 1 4 t 3 2 e 3 t 3 + e 2 t 6 12 u ( t 3) . Since y = x x (see the first equation in the given system), we obtain y ( t ) = 1 2 + 4 e t 3 + e 2 t 6 3 4 t 3 2 2 e 3 t 3 e 2 t 6 12 u ( t 3) . 14. Since L{ x 00 } ( s ) = s 2 X ( s ) sx (0) x (0) = s 2 X ( s ) s, L{ y 00 } ( s ) = s 2 Y ( s ) sy (0) y (0) = s 2 Y ( s ) , applying the Laplace transform to the given equations yields s 2 X ( s ) s = Y ( s ) + e s /s s 2 Y ( s ) = X ( s ) + (1 /s ) e s /s s 2 X ( s ) Y ( s ) = s + ( e s /s ) X ( s ) + s 2 Y ( s ) = (1 /s ) ( e 3 s /s ) . Solving for X ( s ) yields X ( s ) = s 4 + 1 s ( s 4 1) + s 2 1 s ( s 4 1) e s = s 4 + 1 s ( s 4 1) + 1 s ( s 2 + 1) e s = 1 s + 1 2 1 s + 1 + 1 2 1 s 1 + s s 2 + 1 + 1 s s s 2 + 1 e s . Using linearity of the inverse Laplace transform and formula (6) in Section 7.6, we get x ( t ) = 1 + e t 2 + e t 2 + cos t + [1 cos( t 1)] u ( t 1) = cosh t + cos t 1 + [1 cos( t 1)] u ( t 1) . Since y = x 00 u ( t 1) (see the first equation in the system), after some algebra we obtain y ( t ) = cosh t cos t [1 cos( t 1)] u ( t 1) . 16. First, note that the initial conditions are given at the point t = . Thus, for the Laplace transform method, we have to shift the argument to get zero initial point. Let us denote w ( t ) := x ( t + 1) and v ( t ) := y ( t + ) . 256 Exercises 7.9 The chain rule yields w ( t ) = x ( t + )( t + ) = x ( t + ) , v ( t ) = y ( t + )( t + ) = y ( t + ) ....
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 Spring '08
 MAZMANI
 Equations

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