Assignment 34

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course Mth 279

8/6

Query 32 Differential Equations*********************************************

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)

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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:

Graph is of positive humps starting at

L{f(t)} = int(e^(-st)sin(t) + int(0e^(-st))

I’m at a loss of how to finish solving this problem.???

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You're on the right track.

You have to integrate e^(-st) sin(t) only from 0 to pi. The integral of e^(-s t) * 0 from pi to 2 pi is zero.

The result is the integral over the first period of the function.

By Theorem 7.6, the Laplace transform of the entire function is then equal to this result, divided by 1 - e^(-s T) = 1 - e^(-2 pi s).

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Given Solution:

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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:

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

L^-1{(s^2-s)/s^3}

partial fractions

A/s + B/s^2 + C/s^3

s^2-s = As^2 + Bs + C

A=1

B=-1

C=0

F(s) = 1/s -1/s^2

L^(-1){1/s} = 1

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

f(t) = 1-t

L^(-1){e^-s/(s(1-e^-3))} = L^(-1){e^-s/s} * L^(-1){1/(1-e^(-3))}

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The Laplace transform does not distribute over multiplication.

Note that Theorem 7.6 addresses the meaning of 1 - e^(-s) in the denominator.

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L^(-1){e^-s/s} gen Laplace e^(-as)/s

gen f(t) h(t-a)

a=1

f(t) = h(t-1)

L^(-1){1/(1-e^-3)}

I am not sure how to take this L^(-1)????

f(t) = 1 - t + h(t-1) + L^(-1){1/(1-e^(-3))}

I am just not sure about the last L^(-1) other than that I feel ok about this question.

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Given Solution:

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Self-critique (if necessary):

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