Query_31

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course Mth 279

5/24 3:33am only one more to type!

Query 31 Differential Equations

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Question: Using, if necessary, the table in your text, find the Laplace transform of e^(3 t - 3) * h(t - 1), where h(t) is the Heaviside function.

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Your solution:

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

From the table

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

and

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

The solution may look like

e^(-s)/(s-3)

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

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From the table you know that the transform of

H(t-c) * f(t - c) = e^(-c s) F(s).

So the result is

e^(-s) F(s) = e^(-s) * 1 / (s - 3) = e^(-s) / (s - 3).

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It is instructive to actually do the integral.

h(t-1) is zero up to t = 1, then has value 1.

So h(t-1) e^(3 t - 3) is zero up to t = 1, after which its value is that of e^(3 t - 3).

So integrating

e^(-st) h(t-1) e^(3 t - 3) from t = 0 to infinity,

we get 0 from t = 0 to 1, plus the integral of e^(-st) e^(3 t - 3) from t = 1 to infinity.

e^(-st) e^(3 t - 3) = e^ (3 ( t - 1) - s t)

= e^((3 - s) t - 3)

with antiderivative 1 / (3 - s) e^( (3 - s) t - 3).

For s > 3, this antiderivative approaches 0 as t approaches infinity.

At t = 1 the antiderivative is

1 / (3 - s) e^( (3 - s) * 1 - 3) = e^-s / (3 - s).

Thus the integral is

0 - e^(-s) / (3 - s)

= e^(-2) / (s - 3), s > 3.

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confidence rating #$&*:

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2

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Question: Using, if necessary, the table in your text, find the Laplace transform of e^(2 t) cos(3 t).

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Your solution:

From the table L{e^(a t) cos(omega t)} = (s-a)/((s-a)^2 + omega^2)

L{e^(2 t) cos(3 t)} = (s-2)/((s-2)^2 + 3^2) = (s-2)/((s-2)^2 + 9)

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3

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Given Solution:

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Question: Using, if necessary, the table in your text, find the inverse Laplace transform of 10 / (s^2 + 25) + 4 / (s - 3).

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Your solution:

Using the table,

2 L{sin(5t)} + 4 L{e^(3t)} or L{2sin(5t) + 4e^(3t)}

Why?

10/(s^2 + 25) /2 = 5/(s^2 + 25)

if omega = 5, then omega^2 = 25

L{sin(omega t)} = omega/(s^2 + omega^2)

And

4 / (s - 3) /4 = 1 / (s - 3)

L{e^(at)} = 1/(s-a)

a = 3

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3

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Given Solution:

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Question: Using, if necessary, the table in your text, find the inverse Laplace transform of e^(-2 s) / (s - 9).

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Your solution:

Using the table, I'm gonna assume here using some properties that I have noticed.

L{e^(9t)h(t-2)}

Why?

L{e^(9t)} = 1/(s - 9)

And

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

confidence rating #$&*:

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3

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Question: Using, if necessary, the table in your text, find the inverse Laplace transform of 1 / (s + 1)^3

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Your solution:

Using the table

L{t^n} = n!/s^(n+1)

+from Query_30

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

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

L{1} = (1/s)

Also found the correct one in the table

L{e^(at)t^n} = n!/(s-a)^(n+1)

a = -1

n + 1 = 3

n = 2

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

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3

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Given Solution:

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Question: Using, if necessary, the table in your text, find the inverse Laplace transform of (2 s - 3) / (s^2 - 3 s + 2).

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Your solution:

(2 s - 3) / (s^2 - 3 s + 2) = (2 s - 3) /((s-2)(s-1)) = A/(s-2) + B/(s-1)

(A(s-1) + B(s-2))/((s-2)(s-1)) = (2 s - 3) /((s-2)(s-1))

(As - A + Bs - 2B) = (2 s - 3)

A + B = 2

-A - 2B = -3

Adding these equations solves for B

B =1

A + 1 = 2

A =1

(2 s - 3) /((s-2)(s-1)) = 1/(s-2) + 1/(s-1)

L{e^(2t)+e^t} due to L{e^(at)} = 1/(s-a)

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3

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Given Solution:

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&#Good responses. See my notes and let me know if you have questions. &#