Query_30

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course Mth 279

5/24 3:08 am2 more type up, I'm almost done!

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{f(t)} = Int(0 to infinity) e^(-st) f(t) dt

L{te^(t*sqrt(t))} = Int(0 to infinity) e^(-st)te^(t*sqrt(t)) dt

= Int(0 to infinity) te^(-st + t*sqrt(t)) dt

= Int(0 to infinity) te^(-t (s - sqrt(t)) dt = y

Integration by parts

Int(udv) = uv - Int(vdu)

u = t

du = 1

dv = e^(-ts + t^(3/2))

v = Int (e^(-ts + t^(3/2))) = DNE

Seting e as u, and t as dv would just continue to make the problem more

complicated, so L{te^(t*sqrt(t))} DNE (does not exist)

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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:

L{f(t)} = Int(0 to infinity) e^(-st) f(t) dt

L{0} = Int(0 to infinity) 0e^(-st) dt = [1](0 to infinity) = 1

L{t - 1} =

@&

f(t) is zero until t = 1, so the interval from t = 0 to t = 1 contributes nothing to the integral.

The correct integral would be

Int(1 to infinity) e^(-st)(t - 1) dt.

So where you have substituted 0 for your antiderivative, you'll want to substitute 1.

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= Int(0 to infinity) te^(-st) dt - Int(0 to infinity) e^(-st) dt

Integration by parts -> Int(udv) = uv - Int(vdu)

u = t

du = 1

dv = e^(-st)

v = (-1/s)e^(-st)

= (-t/s)e^(-st) + (1/s)Int(0 to infinity)e^(-st) dt - Int(0 to infinity)e^(-st) dt

= (-t/s)e^(-st) - (1/s^2)e^(-st) + (1/s)e^(-st)

Now to evaluate from zero to infinity

= (-t/s)e^(-st)|_infinity + (-0/s)e^(-s*0) - (1/s^2)e^(-st)|_infinity + (1/s^2)e^(-s*0) + (1/s)e^(-st)|_infinity - (1/s)e^(-s*0)

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

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Given Solution:

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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:

f(t) = cos(omega t)

L{cos(omega t)} = Int(0 to infinity) e^(-st) cos(omega t) dt = y

Integration by parts -> Int(udv) = uv - Int(vdu)

u = cos(omega t)

du = -omega*sin(omega t)

dv = e^(-st)

v = (-1/s)e^(-st)

y = cos(omega t)(-1/s)e^(-st) - Int (-1/s)e^(-st)(-omega*sin(omega t))dt

= (-e^(-st)/s)cos(omega t) - (omega/s) Int e^(-st)sin(omega t)dt

Integration by parts -> Int(udv) = uv - Int(vdu)

u = sin(omega t)

du = omega*cos(omega t)

dv = e^(-st)

v = (-1/s)e^(-st)

y = (-e^(-st)/s)cos(omega t) - (omega/s)[ (-e^(-st)/s)sin(omega t) + (omega/s) Int e^(-st)cos(omega t)dt]

y + (omega^2/s^2)y = (-e^(-st)/s)cos(omega t) + (-0mega*e^(-st)/s^2)sin(omega t)

((s^2 + omega^2)/s^2)y = -e^(-st)[(1/s)cos(omega t) - (0mega/s^2)sin(omega t)]

Evaluating the right side from zero to infinity.

0 - (-1/s)cos(0) + 0 - (omega/s^2)sin(0)

((s^2 + omega^2)/s^2)y = (1/s)

y = (1/s) * (s^2 /(s^2 + omega^2))

= s/(s^2 + omega^2)

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Given Solution:

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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:

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

L{e^(3 t) sin(t)} = Int(0 to infinity) e^(-st) e^(3 t) sin(t) dt

= Int(0 to infinity) e^(-(s-3)t)sin(t) dt = y

Integration by parts -> Int(udv) = uv - Int(vdu)

u = sint

du = cost

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

v = (-1/(s-3))e^(-(s-3)t)

y = -sint(e^(-(s-3)t)/(s-3)) + (1/(s-3)) Int e^(-(s-3)t)cos(t) dt

Integration by parts -> Int(udv) = uv - Int(vdu)

u = cost

du = -sint

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

v = (-1/(s-3))e^(-(s-3)t)

y = -sint(e^(-(s-3)t)/(s-3)) + (1/(s-3)) [(-1/(s-3))cost e^(-(s-3)t) - (1/(s-3)) Int e^(-(s-3)t)sin(t) dt]

y + (1/(s-3)^2)y

(((s-3)^2 + 1)/(s-3)^2)y = -sint(e^(-(s-3)t)/(s-3)) - (1/(s-3)^2)cost e^(-(s-3)t)

Evalutae right side from zero to infinity. s>3

sin(0)(e^0/(s-3)) + (1/(s-3)^2)cos(0) e^0

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

y = (1/(s-3)^2) * (((s-3)^2)/(s-3)^2 + 1))

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

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Given Solution:

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Self-critique (if necessary):

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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. &#