Consider the following initial value problem, representing the response of a damped oscillator subject to the discontinuous applied fo (1 2

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I understand how to find the laplace transform of f(t) but i dont understand exactly what the questions are asking. This is practice 

Consider the following initial value problem, representing the response of a damped oscillator subject to the discontinuous applied force f(t):
{o
Si 2<t< 6,
y' + 10y + 29y = f(t), y(0) = -6,
y(0) = 2, f(t) =
otherwise.
In the following parts, use h(t – c) for the Heaviside function he(t) when necessary.
a. First, compute the Laplace transform of f(t).
L {f(t)}(s) =
b. Next, take the Laplace transform of the left-hand-side of the differential equation, set it equal to your answer from Part a. and solve for L{y(t)}.
L {y(t)}(s) =
c. We will need to take the inverse Laplace transform in order to find y(t). To do so, let's first rewrite L{y(t)} as
(; - ro).
1
1
L {y(t)}(s) =
29
6s
e
- 6F(s) + 2G(s)
2s
F(s)
F(s)
S
29
where
F(s) =
and G(s) =
d. Part c. indicates we will need L-1{F(s)} and L-1{G(s)}. Let's go ahead and find those now.
L-1{F(s)}(t) =
and L-1{G(s)}(t) =
e. Use your answer in Part d. to compute L-1 {e¯2° F(s)} and L-1 {e_6$F(s)}.
L-1{ =
e-2° F(s)}(t)
and L-1{e-6*F(s)}(t) =
f. Finally, combine all the previous steps to write down y(t).
Transcribed Image Text:Consider the following initial value problem, representing the response of a damped oscillator subject to the discontinuous applied force f(t): {o Si 2<t< 6, y' + 10y + 29y = f(t), y(0) = -6, y(0) = 2, f(t) = otherwise. In the following parts, use h(t – c) for the Heaviside function he(t) when necessary. a. First, compute the Laplace transform of f(t). L {f(t)}(s) = b. Next, take the Laplace transform of the left-hand-side of the differential equation, set it equal to your answer from Part a. and solve for L{y(t)}. L {y(t)}(s) = c. We will need to take the inverse Laplace transform in order to find y(t). To do so, let's first rewrite L{y(t)} as (; - ro). 1 1 L {y(t)}(s) = 29 6s e - 6F(s) + 2G(s) 2s F(s) F(s) S 29 where F(s) = and G(s) = d. Part c. indicates we will need L-1{F(s)} and L-1{G(s)}. Let's go ahead and find those now. L-1{F(s)}(t) = and L-1{G(s)}(t) = e. Use your answer in Part d. to compute L-1 {e¯2° F(s)} and L-1 {e_6$F(s)}. L-1{ = e-2° F(s)}(t) and L-1{e-6*F(s)}(t) = f. Finally, combine all the previous steps to write down y(t).
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