#$&*
course Mth 279
5/24 3:49 amLast one, I can take my test as soon as I read over my errors and the problems that I could not figure out. Can't Wait!
Query 32 Differential Equations
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Question: Use Laplace transforms to solve the equation y ' + 2 y = 4 t, with initial condition y(0) = 3. Verify your solution.
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Your solution:
Y(s) = L{y(t)}
L{y'} + 2 L{y} = 4/s^2
L{y'} = sY(s) - y(0) = sY(0) - 3
sY(s) + 2Y(s) - 3 = 4/s^2
sY(s) + 2Y(s) = 4/s^2 + 3
Y(s)(s + 2) = (3s^2 + 4)/s^2
Y(s) = (3s^2 + 4)/(s^2(s + 2))
(3s^2 + 4)/(s^2(s + 2)) = A/s^2 + B/s + C/(s+2)
A(s+2) + Bs(s+2) + Cs^2 = 3s^2 + 4
As + 2A + Bs^2 + 2Bs + Cs^2 = 3s^2 + 4
B+C = 3
A + 2B = 0
2A = 4
A = 2
B = -1
C = 4
Y(s) = 2/s^2 - 1/s + 4/(s+2)
y(t) = 2 L^(-1){1/s^2} - L^(-1){1/s} + 4 L^(-1){1/(s+2)}
= 2t - 1 + 4e^(-2t)
@&
Good.
You could also inverse transform
(3s^2 + 4)/(s^2(s + 2))
by splitting it into
(3s^2)/(s^2(s + 2)) = 1 / (s + 2)
and
4 /(s^2(s + 2))
The latter would still require partial fractions with 3 parameters so there's not much advantage to doing this, but this sort of rearrangement can at times be useful.
*@
Step by step solution for verification:
u(t) = e^(2t)
e^(2t)y' + 2e^(2t)y = 4te^(2t)
(e^(2t)y(t))' = 4te^(2t)
Integration of both sides yeilds:
e^(2t)y(t) = 4e^(2t)(t/2 - 1/4) + c
divide by u(t)
y(t) = 4(t/2 - 1/4) + c/e^(2t) = 2t + c/e^(2t) - 1
y(0) = 3
y(0) = 2*0 + ce^(-2*0) - 1
c - 1 = 3
c = 4
y(t) = 4e^(-2t) + 2t - 1
Which is the same answer to the previous problem.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Question: Let f(t) = sin(t) for 0 <= t < pi, f(t) = 0 for pi <= t < 2 pi, with f(t + 2 pi) = f(t).
Graph this periodic function and find its Laplace transform.
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Your solution:
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The graph of this function just consists of the positive 'humps' of the sine function.
The Laplace transform of this function over its first period is
integral ( e^(-s t) sin(t), t from 0 to pi ) + integral(0, t from pi to 2 pi) = ( -e^(-pi s) - e^(-pi * 0)) / (s^2 + 1) = (1 - e^(-pi s) ) / (s^2 + 1).
The Laplace transform of a function of period T is 1 / (1 - e^(-sT) ) times the Laplace transform of that function restricted to the interval 0 <= t <= T, so the transform of this function is
The period of this function is T = 2 pi so its Laplace Transform is
F(s) = ( (1 - e^(-pi s) ) / (s^2 + 1) ) * 1 / (1 - e^(-2 pi s)
= 1 / ( (s^2 + 1)(1 + e^(pi s) ).
Note that 1 - e^(-2 pi s) = 1 - e^(-pi s)^2 = (1 - e^(-pi s) ) ( 1 + e^(-pi s) ), which leads to the last step of the simplification.
Compare this with the Laplace transform of sin(t), which is 1 / (s^2 + 1).
Eliminating the half-cycle where sin(t) is negative reduces the Laplace Transform by factor 1 + e^(-pi s).
*@
qq
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Question: Find the function whose Laplace transform is (s^2 - s) / s^3 + e^(-s) / (s ( 1 - e^(-s) ).
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Your solution:
Only half appears to exist, as could not find a form or do partical fractions corectly for the second part.
maybe L^(-1){e^(-s) / (s ( 1 - e^(-s) )} simply DNE
@&
One thing we all learn is that just because we can't find it doesn't mean it doesn't exist.
Sometimes things exist but they simply cannot be put into closed form.
Other times they exist but we're just not up to the task of finding them. That's what keeps mathematicians busy.
*@
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1/s is the Laplace transform of the function
f(t) = 1.
If the transform of f(t) is F(s) then the transform of
h(t-c) * f(t)
is just e^(-s) F(s).
So e^(-s) / s would be the transform of
h(t-1).
Dividing by (1 - e^(-s)) indicates a periodic function with period 1.
*@
(s^2 - s) / s^3 = A/s^3 + B/s^2 + C/s
A + Bs + Cs^2 = s^2 - s
A = 0
B = -1
C = 1
(s^2 - s) / s^3 = -1/s^2 + 1/s
L^(-1){-1/s^2+ 1/s} = L^(-1){-1/s^2} + L^(-1){1/s} = -t + 1
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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#*&!
&#Good responses. See my notes and let me know if you have questions. &#