Query 30

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course MTH 279

Query 30 Differential Equations*********************************************

Question: Using the definition of the Laplace transform, find the Laplace transform of f(t) = t e^(t sqrt(t)).

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Your solution:

F(s) = ʃte^(t^(3/2))e^(-st)dt, 0 to ∞ = lim(ʃt*e^(t(t^(1/2)-s))dt, 0 to T) as T goes to ∞. This will not work, because for any fixed value of s, the function diverges whenever t>s.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

As far as I know, my reasoning is sound. The limit of e^(t) goes to ∞ when t>0, and will be complex if t<0. So if s is kept constant, there are no values that lead to a convergent exponential, and if t is 0, F(s) would be 0 as well.

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Self-critique rating: 3

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I agree. There is no value of s for which e^(t^(3/2) - s t) does not approach infinity.

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Question: Using the definition of the Laplace transform, find the Laplace transform of the function f(t) defined by f(t) = 0, 0 <= 1 < 1; f(t) = t - 1, 1 <= t.

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Your solution:

To express the bounded integral, we’ll have ʃf(t)dt{x,y} to mean the integral of f(t) with respect to dt from x to y. Any time there are brackets {}, it is to indicate bounds of integration.

After examining the function, we determine that f(t)=0 will contribute nothing to our integrals, so we can essentially “leave it out.” We will integrate on the interval t>=1, with f(t)=t-1.

That makes our Laplace Transform F(s) = ʃ(t-1)e^(-st)dt{1,∞}. This can be rewritten as F(s) = ʃte^(-st)dt{1,∞} - ʃe^(-st)dt{1,∞}.

We use integration by parts to get F(s) = -te^(-st)/s{1,∞} - ʃ-e^(-st)/sdt{1,∞} - ʃe^(-st)dt{1,∞}.

This all simplifies by algebraic manipulation down to -e^(-st)(st+1)/s^2{1,∞} + e^(-st)/s{1,∞}. Using our bounds, and simplifying, we get F(s) = e^(-s)/s^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 the definition of the Laplace transform, find the Laplace transform of f(t) = cos(omega t).

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Your solution:

F(s) = ʃcos(ωt)e^(-st)dt{0,∞}.

We do integration by parts and algebraic manipulation to get the form ʃcos(ωt)e^(-st)dt =[-s*cos(ωt)e^(-st)+ωsin(ωt)e^(-st)]/(ω^2+s^2).

Using this integration, we now have F(s) = [(-s*cos(ωt)e^(-st)+ωsin(ωt)e^(-st))/(ω^2+s^2)]{0,∞}. When we evaluate this using our integration bounds, we get:

F(s) = 0 - (-s/(ω^2+s^2)) = s/(ω^2+s^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 the definition of the Laplace transform, find the Laplace transform of f(t) = e^(3 t) sin(t).

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Your solution:

F(s) = ʃe^(3t)sin(t)e^(-st)dt{0,∞} = ʃe^((3-s)t)sin(t)dt{0,∞}. α=3-s.

F(s) = ʃe^(αt)sin(t)dt{0,∞}.

Using integration by parts, and algebraic manipulation, we get:

F(s) = [-e^(αt)cos(t)+αe^(αt)sin(t)]/(1+α^2){0,∞}. By inspection, we know that in order for the limit of the improper integral to converge, α must be negative. If α=3-s, we know that s>3 in order for the limit to converge. With this in mind, we rewrite as so:

F(s) = [-e^(-(s-3)t)cos(t) + -(s-3)e^(-(s-3)t)sin(t)]/(1+(-(s-3))^2){0,∞}. Solving at our bounds of integration, we find: F(s) = 0 - (-1/(1+(s-3)^2) = 1/(1+(s-3)^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 "

Self-critique (if necessary):

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Self-critique (if necessary):

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#*&!

&#Good work. Let me know if you have questions. &#