Question

Solve the IVP d^2y/dt^2 + 2dy/dt + 17y = 0, y(0) = 0, y'(0) = 6 The Laplace transform of the solutions is L{y} = 6/(s^2+2s+17) By completing the square in the denominator we see that this is the Laplace transform of shifted by the rule (Your first answer blank for this question should be a function of t). Therefore the solution is y =

          Solve the IVP
d^2y/dt^2 + 2dy/dt + 17y = 0, y(0) = 0, y'(0) = 6
The Laplace transform of the solutions is L{y} = 6/(s^2+2s+17)
By completing the square in the denominator we see that this is the Laplace transform of shifted by the rule
(Your first answer blank for this question should be a function of t).
Therefore the solution is y =

Solve the IVP
d^2y/dt^2 + 2dy/dt + 17y = 0, y(0) = 0, y'(0) = 6
The Laplace transform of the solutions is Ly = 6/(s^2+2s+17)
By completing the square in the denominator we see that this is the Laplace transform of shifted by the rule
(Your first answer blank for this question should be a function of t).
Therefore the solution is y =

Added by Melissa P.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Zhumagali Shomanov

05/26/2023

Step 1

To find the solution in the time domain, we need to take the inverse Laplace transform of L{y}. Completing the square in the denominator, we get s^2 + 2s + 17 = (s + 1)^2 + 16. So, L{y} = 6/[(s + 1)^2 + 16]. Using the Laplace transform table, we know that Show more…

Show all steps

Thanks for your feedback!

Solve the IVP The Laplace transform of the solutions is L{y} = 6/s^2 + 2s + 17. By completing the square in the denominator, we see that this is the Laplace transform of shifted by the rule. (Your first answer blank for this question should be a function of t). Therefore, the solution is y =