Assignment 32

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course Mth 279

8/6

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:

L{f(t)} = int(te^(tsqt(t))*e^(-st) from 0 to infinity

= lim T->infinity int(te^(tsqt(t))*e^(-st)dt from 0 to T

lim int(te^(tsqt(t)-st))

???

Unsure how to go about integral???

confidence rating #$&*:

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Given Solution:

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As it turns out 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 <= t < 1; f(t) = t - 1, 1 <= t.

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Your solution:

lim int(t-1)e^(-st))dt

u = t-1

du = dt

dv= e^(-2t)

v = -1/s e^(-st)

-(t-1)e^(-2t)/s ]1toT - int(-1/2 e^(-st))dt

L{f(t)} = e^-s/s^2

confidence rating #$&*:

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1

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Given Solution:

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Self-critique (if necessary):

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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:

using int(e^(au)cos(bu) = e^(au)(acos(bu)+bsin(bu))/(a^2+b^2)

lim{e^(-st)(-scos(wT) + wsin(wT))/(s^2+w^2)) - e^(-s(0))(-scos(w(0))+wsin(w(0))/(s^2+w^2))

0 + s/(s^2+w^2)

L{f(t)} = s/(s^2+w^2)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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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:

using int(e^(au)sin(bu) = e^(au)(asin(bu) - bcos(bu))/(a^2+b^2)

= lim {e^(t(3-s))((3-2)sin(t)-cos(t))/((3-s)^2+1)) from 0 to T

e^(T(3-s))((3-2)sin(T)-cos(T))/((3-s)^2+1)) - e^0((3-s)sin(0) - cos(0))/(3-s)^2 + 1)

0 - -1/((3-s)^2 + 1)

L{f(t)} = 1/((3-s)^2 + 1)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

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&#This looks good. See my notes. Let me know if you have any questions. &#