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nagle_differential_equations_ISM_Part45

nagle_differential_equations_ISM_Part45 - Chapter 7 6...

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Chapter 7 6. Applying the Laplace transform to both sides of the given equation and using Theorem 5 in Section 7.3 to express L { y } and L { y } in terms of Y , we obtain ( s 2 Y - 2 s - 7 ) - 4 ( sY - 2) + 5 Y = 4 s - 3 Y = 1 s 2 - 4 s + 5 2 s - 1 + 4 s - 3 = 2 s 2 - 7 s + 7 ( s - 3) [( s - 2) 2 + 1 2 ] = 2 s - 3 + 1 ( s - 2) 2 + 1 2 . Taking now the inverse Laplace transform and using its linearity and Table 7.1 from Section 7.2 yields y ( t ) = L - 1 2 s - 3 + 1 ( s - 2) 2 + 1 2 ( t ) = 2 e 3 t + e 2 t sin t . 8. Applying the Laplace transform to both sides of the given equation and using Theorem 5 in Section 7.3 to express L { y } and L { y } in terms of Y , we obtain ( s 2 Y - 3 ) + 4 Y = 8 s 3 - 4 s 2 + 10 s Y = 1 s 2 + 4 3 + 8 s 3 - 4 s 2 + 10 s = 3 s 3 + 10 s 2 - 4 s + 8 s 3 ( s 2 + 2 2 ) = 2 s 3 - 1 s 2 + 2 s - 2 s s 2 + 2 2 + 2 2 s 2 + 2 2 . Taking now the inverse Laplace transform and using its linearity and Table 7.1 from Section 7.2 yields y ( t ) = L - 1 2 s 3 - 1 s 2 + 2 s - 2 s s 2 + 2 2 + 2 2 s 2 + 2 2 ( t ) = t 2 - t + 2 - 2 cos 2 t + 2 sin 2 t . 10. Applying the Laplace transform to both sides of the given equation and using Theorem 5 in Section 7.3 to express L { y } and L { y } in terms of Y , we obtain ( s 2 Y - 5 ) - 4 Y = 4 s 2 - 8 s + 2 Y = 1 s 2 - 4 5 + 4 s 2 - 8 s + 2 = 5 s 3 + 2 s 2 + 4 s + 8 s 2 ( s + 2) 2 ( s - 2) = 2 ( s + 2) 2 - 1 s + 2 - 1 s 2 + 1 s - 2 . Taking now the inverse Laplace transform and using its linearity and Table 7.1 from Section 7.2 yields y ( t ) = L - 1 2 ( s + 2) 2 - 1 s + 2 - 1 s 2 + 1 s - 2 ( t ) = 2 te - 2 t - e - 2 t - t + e 2 t . 216

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Exercises 7.5 12. Since the Laplace transform approach requires that initial conditions are given at the origin, we make a shift in argument. Namely, let y ( t ) := w ( t - 1). Then y ( t ) = w ( t - 1)( t - 1) = w ( t - 1) , y ( t ) = w ( t - 1)( t - 1) =
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