Find L1 1s2 4s by 3 ways by the first translation theorem b

Find L^1 {1/s^2 + 4s} by 3 ways: by the first translation theorem by integral transform by convolution theorem

Solution

Translation Theorem:

L^-1 {1/s²+4s} = L^-1 {1/s²+4s + 4 - 4}

= L^-1 {1/(s+2)² - 4}

= 1/2 L^-1 {2/(s+2)² - 2²}

= 1/2 L^-1 {2/s² - 2²}_{s+2 --> s}

= 1/2 e^-2t sin(h)(2t)

= 1/2 e^-2t {e^2t - e^-2t/2}

= 1 - e^-4t /4

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Integral transform (by LaPlace):

L^-1 {1/s²+4s} = L^-1 {1/s(s+4)}

= L^-1 {1/4s - 1/4(s+4)}

= L^-1 {1/4s} - L^-1 {1/4(s+4)}

= 1/4 L^-1 {1/s} - 1/4 L^-1 1/4(s+4)}

= 1/4(1) -1/4 e^-4t

= 1 - e^-4t / 4

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Convolution theorem:

L^-1 {1/s²+4s} = L^-1 {1/s(s+4)}

taking function seperately;

f(s) = 1/s

f´(s) = 1/s+4

then we have;

L{1} = 1/s

f(s) = 1

and

L{e^-4t} = 1/s+4

f´(s) = e^-4t

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Letting f´´(s) be the product of f(s)f´(s):

L {f(t)} = f´´(s)

f(t) = L^-1 {f´´(s)}

= L^-1 {L(f x f´)(t)}

= (f x f´)(t)

= ₀^t f(t-u)f´(u) du

= ₀^t 1.e^-4u du

= [e^-4u / -4] |_o^t

= [(e^-4t/-4) - (e⁰/-4)]

= 1 - e^-4t / 4

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Hope it helps!!