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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.
 Spring '08
 MAZMANI
 Equations

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