Assignment 33

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course Mth 279

8/6

Query 31 Differential Equations*********************************************

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:

f(t) = e^(3 t - 3) * h(t - 1)

in form of h(t-a)f(t-a) = e^(-as)F(s)

a_1=1 from h(t-1) = e^-s

e^(at) = 1/(s-a) a_2=3

1/(s-3)

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

confidence rating #$&*:

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Given Solution:

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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:

f(t) = e^(2 t) cos(3 t)

gen form

a^(at)cos(wt) = (s-a)/((s-a)^2 + w^2)

a= 2 w=3

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

confidence rating #$&*:

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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:

Kind of working backwards

F(s) = 10 / (s^2 + 25) + 4 / (s - 3)

find L^(-1){F(s)}

take each part of function separately

10/(s^2+25)

Laplace form of -> w/(s^2+w^2)

Function ->f(t) =sin(wt)

2(5/(s^2+5^2) w=5

f(t) = 2sin(5t)

4/(s-3)

Laplace form 1/(s-a)

Function f(t)=e^(at)

a=3

4(1/(s-3))

f(t) = 4e^(3t)

f(t) = 2sin(5t)+4e^(3t)

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

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Your solution:

L{f(t)} = e^(-2 s) / (s - 9) = e^(-2s)(1/(s-9))

e^(-2s) -> Laplace e^(-as)F(s)

gen f(t) -> f(t-a)h(t-a)

f(t) = f(t-2)h(t-2)

1/(s-9) -> gen Laplace 1/(s-a) a=0

gen f(t) = e^(at)

f(t) = e^(9t)

f(t)= e^(9(t-2))h(t-2) = L^(-1){f(t)}

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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:

F(s) = 1/(s+1)^3

gen Laplace - n!/(s-a)^(n+1) n=2 a=-1

gen f(t) => a^(at)t^n

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

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

L^-1{F(s)} = 1/2e^(-t)t^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, 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:

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

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

A+B=2

-2A-B=-3

A=1

B=1

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

1/(s-1)

gen Laplace -> 1/(s-a)

f(t) = e^(at) a_1=1

a_2 = 2

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

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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&#Very good work. Let me know if you have questions. &#