Question

(6 points) Problem 5: Find the Inverse Laplace Transform for each of the following: \begin{align} a) \ F(s) &= \frac{3s - 7}{s^2 + 16} \\ b) \ G(s) &= \frac{1 - 3s}{s^2 + 2s + 10} \end{align} Hint: $\sin(at) \xrightarrow{Laplace} \frac{a}{s^2 + a^2}$, $\cos(at) \xrightarrow{Laplace} \frac{s}{s^2 + a^2}$ (5 points) Problem 6: Solve the initial value problem by using Laplace transform: $y'' - y' - 2y = e^t$ $y(0) = 0, y'(0) = 1$

          (6 points) Problem 5: Find the Inverse Laplace Transform for each of the following:
\begin{align*}
a) \ F(s) &= \frac{3s - 7}{s^2 + 16} \\
b) \ G(s) &= \frac{1 - 3s}{s^2 + 2s + 10}
\end{align*}
Hint: $\sin(at) \xrightarrow{Laplace} \frac{a}{s^2 + a^2}$, $\cos(at) \xrightarrow{Laplace} \frac{s}{s^2 + a^2}$
(5 points) Problem 6: Solve the initial value problem by using Laplace transform:
$y'' - y' - 2y = e^t$
$y(0) = 0, y'(0) = 1$

(6 points) Problem 5: Find the Inverse Laplace Transform for each of the following:

a) F(s) = (3s - 7)/(s^2 + 16)

b) G(s) = (1 - 3s)/(s^2 + 2s + 10)

Hint: sin(at) Laplace(a)/(s^2 + a^2), cos(at) Laplace(s)/(s^2 + a^2)
(5 points) Problem 6: Solve the initial value problem by using Laplace transform:
y” - y' - 2y = e^t
y(0) = 0, y'(0) = 1

Added by Adri-N W.

Advanced Engineering Mathematics

Erwin Kreyszig 9th Edition

Chapter 6

Instant Answer

Solved by Expert Shu-Ting Huang

Step 1

First, we factor the denominator: 1-3s = -(3s-1). Then, we can write the expression as: (s+16)/(1-3s) = A/(3s-1) + B To find A and B, we can multiply both sides by 1-3s and equate the coefficients of the terms on both sides: s+16 = A + B(3s-1) Expanding and Show more…

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Problem 5: Find the inverse Laplace Transform for each of the following: a) L^(-1){(s+16)/(1-3s)} b) L^(-1){(s+2)/(s+10)} Problem 6: Solve the initial value problem by using Laplace transform: y'' - 2y' + y = y(0) = 0, y'(0) = 1.