nagle_differential_equations_ISM_Part45

nagle_differential_equations_ISM_Part45 - Chapter 7 6....

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Unformatted text preview: 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 00 } 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 00 } 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 00 } 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 ....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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nagle_differential_equations_ISM_Part45 - Chapter 7 6....

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