Question

For problems 6, 7, and 8, use a Laplace transform table and the linearity property of the Laplace transform to determine the following transforms: 6. L{6e^-3t - t^2 + 2t - 8} 7. L{5 - e^2t + 6t^2} 8. L{t^3 - te^t + e^4t cos(t)} 9. L{t^2 + e^t sin 2t} 10. L{e^-t cos 3t + e^6t - 1} 11. L{(t - 1)^4} 12. L{cosh(bt)} 13. L{cos^3 t} 14. L{(1 + e^-t)^2} 15. L{te^2t cos 5t}

          For problems 6, 7, and 8, use a Laplace transform table and the linearity property of the Laplace transform to determine the following transforms:
6. L{6e^-3t - t^2 + 2t - 8}
7. L{5 - e^2t + 6t^2}
8. L{t^3 - te^t + e^4t cos(t)}
9. L{t^2 + e^t sin 2t}
10. L{e^-t cos 3t + e^6t - 1}
11. L{(t - 1)^4}
12. L{cosh(bt)}
13. L{cos^3 t}
14. L{(1 + e^-t)^2}
15. L{te^2t cos 5t}

For problems 6, 7, and 8, use a Laplace transform table and the linearity property of the Laplace transform to determine the following transforms:
6. L6e^-3t - t^2 + 2t - 8
7. L5 - e^2t + 6t^2
8. Lt^3 - te^t + e^4t cos(t)
9. Lt^2 + e^t sin 2t
10. Le^-t cos 3t + e^6t - 1
11. L(t - 1)^4
12. Lcosh(bt)
13. Lcos^3 t
14. L(1 + e^-t)^2
15. Lte^2t cos 5t

Added by Seth M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

07/27/2023

Step 1

L{6e^{-3t}t^2 + 2t - 8} = 6L{e^{-3t}t^2} + 2L{t} - 8L{1} Now, we can use the Laplace transform table to find the Laplace transforms of each term: L{e^{-3t}t^2} = \frac{2!}{(s+3)^3} = \frac{2}{(s+3)^3} Show more…

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For problems 6, 7, and 8, use the Laplace transform table and the linearity property of the Laplace transform to determine the following transforms: 9{6e-3t t^2 + 2t - 8} 7. 9{5 - e^2t + 6t^2} P{t - e^t + cos(t)} 9. 9{t^2 + e^t sin(2t)} 10. P{e^-t cos(3t) + e^6t - 1} 11. S{(t - 1)'} 12. ~${cosh(bt)} 13. 9{cos(t)} 14. 2{(1 + e^-t)^2} 15. {te^zt cos(5t)}