#$&*
course Mth 279
5/8 2
Query 30 Differential Equations
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Question: Using the definition of the Laplace transform, find the Laplace transform of f(t) = t e^(t sqrt(t)).
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Your solution:
L { t e^(t sqrt(t))} = Int(t e^(t sqrt(t))*e^(-st) dt,0,infinity) = limit as Tinfinty of Int(t e^(t^(3/2)-st) dt)
=lim Tinfin
I’m not sure what I’m doing wrong but I’ve spent probably 2 hrs. trying to figure this problem out and can’t find an anti-derivative for this integral. I’m sure I missing some sort of transformation that is associated with the Laplace transform but I can’t spend any more time on this problem.
confidence rating #$&*:
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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 the function f(t) defined by f(t) = 0, 0 <= 1 < 1; f(t) = t - 1, 1 <= t.
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Your solution:
???I’m confused wouldn’t f(t) =0, for 0<=1 <1 always make f(t) = 0?????????????????
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
@& The integral of f(t) * e^(- s t) from t = 0 to infinity is broken into
integral ( f(t) e^(-s t), t from 0 to 1) + integral( f(t) e^(-st)), t from 1 to infinity).
The first integral is zero, since between t = 0 and t = 1 we have f(t) = 0.
For the second integral f(t) = 1 for all t from 1 to infinity, so we get
integral ( e(-s t), t from 1 to infinity)
Our antiderivative is -1/s e^(-s t).
Between t = 1 and infinity the antiderivative changes from
-1/s * e^(-s) to zero, so the change in the antiderivative is 1/s e^(-s).
This is the Laplace transform of the given function.*@
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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) = cos(omega t).
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Your solution:
L {cos(`omega*t)} = Int(cos(`omega*t)e^(-st) dt,0,infinity)
= limit as Tinfinty of Int(cos(`omega*t)e^(-st) dt)
Using integration by parts we find Int(cos(`omega*t)e^(-st) dt)
v’ = e^(-st), v= (1/-s)e^(-st), u = cos(`omega*t), u’ = omega sin (`omega*t)
= [cos(`omega*t), * (1/-s)e^(-st)]- Int(-se^(-st)* omega sin (`omega*t) dt)
Using integration by parts again
v’ = (1/-s)e^(-st), v =(1/-s^2)e^(-st) u= omega sin (`omega*t), u’ =(-omega^2) cos(`omega*t)
=[ cos(`omega*t), * (1/-s)e^(-st)] - [`omega sin(`omega*t) * (1/-s^2)e^(-st)] -
Int((1/-s^2)e^(-st) *(-omega^2) cos(`omega*t) dt)
We see that we have
[ cos(`omega*t), * (1/-s)e^(-st)] - [`omega sin(`omega*t) * (1/-s^2)e^(-st)]
And our original equation, Int(cos(`omega*t)e^(-st)) which is mult by `-omega^2/s^2
So our function is
F(s)=[ cos(`omega*t), * (1/-s)e^(-st)] - [`omega sin(`omega*t) * (1/-s^2)e^(-st)] - (`-omega^2/s^2)F(s)
= [(e^(-st)/`omega^2)*(`omega sin(`omega t) - s cos(`omega t)] - (s^2/`omega^2)*F(s)
Where evaluated [(e^(-st)/`omega^2)*(`omega sin(`omega t) - s cos(`omega t)] as tinfinity we get
???can’t get it?????
After looking at next question I get it using, Int(e^(alpha*t)*cos(`beta*t) dt) = (e^(`alpha*t))((`alpha cos(`beta t) + `beta sin(beta t)/(alpha^2 + beta^2)
We get
Int(cos(`omega t)e^(-st) dt,0,infinity) = (e^(-st))((-s)cos(`omega t) - `omega sin(`omega t))/( -s^2 + `omega ^2)
Finding limits on integral at 0 and limit as Tinfinty
e^(t(3 -s))*( (3 -s)sin(t) - cos(t))/( s^2 - 6s +10 )
= -s/(( s^2 - `omega^2 ) at lower limit
= 0 for s> 0 as tinfinity
So that
L { = e^(3 t) sin(t)} = s/(( s^2 - `omega^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:
L { = e^(3 t) sin(t)} = Int(e^(3 t) sin(t)*e^(-st) dt,0,infinity)
= Int(e^(t(3 -s)) sin(t) dt,0,infinity)
Knowing….
Int(e^(alpha*t)*sin(`beta*t) dt) = (e^(`alpha*t))((`alpha sin(`beta t) - `beta cos(beta t)/(alpha^2 + beta^2)
We get
Int(e^(t(3 -s)) sin(t) dt,0,infinity) = (e^((3 -s)*t))(( (3 -s)sin(1 t) - 1 cos(1 t))/( (3 -s)^2 + 1^2)
= e^(t(3 -s))*( (3 -s)sin(t) - cos(t))/( s^2 - 6s +10 )
Finding limits on integral at 0 and limit as Tinfinty
e^(t(3 -s))*( (3 -s)sin(t) - cos(t))/( s^2 - 6s +10 )
= -1/(( s^2 - 6s +10 ) at lower limit
= 0 for s> 3
So that
L { = e^(3 t) sin(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:
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@& Good. See my notes on the first questions.*@