nagle_differential_equations_ISM_Part53

# nagle_differential_equations_ISM_Part53 - Chapter 7 Using...

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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 = π . 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 + π ) . 256 Exercises 7.9 The chain rule yields w ( t ) = x ( t + π )( t + π ) = x ( t + π ) , v ( t ) = y ( t + π )( t + π ) = y ( t + π ) ....
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nagle_differential_equations_ISM_Part53 - Chapter 7 Using...

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