Question

Take the Laplace transform of the following initial value and solve for Y(s) = L{y(t)}: y'' + 9y = { sin(???t), 0 ? t < 1; 0, 1 ? t. y(0) = 0, y'(0) = 0 Y(s) = Next take the inverse transform of Y(s) to get y(t) = Use step(t-c) for uc(t) Note: ??? / ((s^2 + ???^2)(s^2 + 9)) = ??? / (???^2 - 9) * (1 / (s^2 + 9) - 1 / (s^2 + ???^2))

          Take the Laplace transform of the following initial value and solve for Y(s) = L{y(t)}:
y'' + 9y = { sin(???t), 0 ? t < 1; 0, 1 ? t. y(0) = 0, y'(0) = 0
Y(s) =
Next take the inverse transform of Y(s) to get
y(t) =
Use step(t-c) for uc(t)
Note: ??? / ((s^2 + ???^2)(s^2 + 9)) = ??? / (???^2 - 9) * (1 / (s^2 + 9) - 1 / (s^2 + ???^2))

Take the Laplace transform of the following initial value and solve for Y(s) = Ly(t):
y” + 9y = sin(???t), 0 ? t < 1; 0, 1 ? t. y(0) = 0, y'(0) = 0
Y(s) =
Next take the inverse transform of Y(s) to get
y(t) =
Use step(t-c) for uc(t)
Note: ??? / ((s^2 + ???^2)(s^2 + 9)) = ??? / (???^2 - 9) * (1 / (s^2 + 9) - 1 / (s^2 + ???^2))

Added by Tracy B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

09/14/2023

Step 1

First, we need to take the Laplace transform of the given equation. The Laplace transform of the second derivative is given by: $$L\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) = s^2Y(s)$$ Since $y(0) = 0$ and $y'(0) = 0$. Show more…

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Take the Laplace transform of the following initial value and solve for Y(s) = L{y(t)}: y'' + 9y = { sin(̀́̂t), 0 <= t < 1; 0, 1 <= t. y(0) = 0, y'(0) = 0. Y(s) = Next, take the inverse transform of Y(s) to get y(t) = Use step(t-c) for uc(t). Note: ̀́̂ / ((s^2 + ̀́̂^2)(s^2 + 9)) = ̀́̂ / (̀́̂^2 - 9) * (1 / (s^2 + 9) - 1 / (s^2 + ̀́̂^2))