Query 31

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course MTH 279

Query 31 Differential Equations*********************************************

Question: Using, if necessary, the table in your text, find the Laplace transform of e^(3 t - 3) * h(t - 1), where h(t) is the Heaviside function.

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Your solution:

F(s) = ʃe^(3t-3)h(t-1)e^(-st)dt{0,∞} = ʃe^(3t-3-st)h(t-1)dt{0,∞} = e^(-3)*ʃe^((3-s)t)h(t-1)dt{0,∞}. We make the substitution that α=3-s, and b=t-1. Since we made the substitution that b=t-1, we modify our bounds and F(s) = e^(-3)ʃe^(αt)h(b)dt{1,∞}. The Heaviside function is 1 at b=1.

We take the integral and get F(s) = e^(-3)*[e^(αt)/α{1,∞}]. Plugging in α, we get:

F(s) = e^(-3)*[e^(-(s-3)t)/-(s-3) {1,∞}]. Evaluating at the bounds, we get:

F(s) = e^(-s)/(s-3).

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Using, if necessary, the table in your text, find the Laplace transform of e^(2 t) cos(3 t).

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Your solution:

F(s) = ʃe^(2t)cos(3t)e^(-st)dt{0,∞}. To save time, I’m going to drop the bounds and add them back at the end, but they are implied from this step to the end.

F(s) = ʃe^(αt)cos(bt)dt; α=2-s, b=3. We know that α<0, so s>2. α = -(s-2).

We use integration by parts to compute the integral, and have F(s) = e^(αt)sin(bt)/(b^2+α^2) + αe^(αt)cos(bt)/(b^2+α^2).

We substituting α and b in, and we get:

F(s) = [e^(-(s-2)t)sin(3t)/(9+(s-2)^2) + -(s-2)e^(-(s-2)t)cos(3t)/(9+(s-2)^2)]{0,∞}.

Evaluating at the bounds, we now have:

F(s) = (s-2)/(9+(s-2)^2).

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Using, if necessary, the table in your text, find the inverse Laplace transform of 10 / (s^2 + 25) + 4 / (s - 3).

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Your solution:

The inverse Laplace, denoted L^(-1)(s) = f(t) = L^(-1)(10/(s^2+25)) + L^(-1)(4/(s-3)).

We look up inverse Laplace transforms and find that the first inverse closely matches that of L(sin(ωt)) = ω/(s^2+ω^2) and the second closely matches L(e^(αt)) = 1/(s-α). Both are off by a constant multiplier (different in each respective case).

Using algebra: f(t) = L^(-1)((5*2)/(s^2+5^2)) + L^(-1)((1*4)/(s-3)). We can draw the constants out, f(t) = 2L^(-1)(5/(s^2+5^2)) + 4L^(-1)(1/(s-3)).

f(t) = 2sin(5t)+4e^(3t).

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I think I did this correctly. It’s hard to mess up tabular answers. The only thing I’m unsure of is my notation. I think it’s self-explanatory, even if not entirely correct, but I believe it is correct, nonetheless.

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Self-critique rating: 3

@&

Your notation is fine, and your final answer does have the given transform. Good solution.

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Question: Using, if necessary, the table in your text, find the inverse Laplace transform of e^(-2 s) / (s - 9).

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Your solution:

This F(s) can be decomposed into L^(-1)(e^(-2s)*1/(s-9)).

This takes the form of e^(-αs)F(s) with α=2. The solution is a shift of the Heaviside function, so L^(-1)(F(s)) = e^(9(t-2))*h(t-2) where h(t-2) is a shift of the Heaviside function.

f(t) = e^(9(t-2))*h(t-2)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Using, if necessary, the table in your text, find the inverse Laplace transform of 1 / (s + 1)^3

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Your solution:

This F(s) takes the form of n!/(s-α)^(n+1) with n = 2, α=-1.

The inverse gives f(t) = ½ * e^(-t)t^2.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Using, if necessary, the table in your text, find the inverse Laplace transform of (2 s - 3) / (s^2 - 3 s + 2).

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Your solution:

This F(s) takes the form of (s-α)/((s-α)^2+ω^2). With some algebraic manipulation, we can say that F(s) = 2(s-3/2)/((s-3/2)^2-(1/2)^2). This means that α = 3/2, and ω=i/2.

f(t) = 2e^(αt)cos(ωt) = 2e^(3t/2)cos(it/2) = e^(3t/2)cosh(t/2).

Another way to do this is to use partial fraction decomposition to get:

F(s) = (2s-3)/(s^2-3s+2) = 1/(s-2) + 1/(s-1).

L^(-1)(F(s)) = e^(2t) + e^(t).

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I actually did the first part of this solution by hand, and at the end when I decided to check my work, I used WolframAlpha to find the inverse laplace for the given F(s), and it gave me the answer in the form e^(t)+e^(2t). I was curious, so I entered my f(t) = e^(3t/2)cosh(t/2) to find the laplace transform for this, and it works out to our given F(s). So this F(s) has two inverse solutions, I guess. I thought it was interesting. Is there any special reason why this is so? Is there any significance to this instance?

@&

cosh(t/2) = (e^(t/2) + e^(-t/2) ) / 2, so when multiplied by e^(3t/2) the result is

(e^(2t) + e^t) / 2

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Self-critique rating: 3 "

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#