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lions [1.4K]
2 years ago
14

Inveres laplace transform (3s-14)/s^2-4s+8​

Mathematics
1 answer:
Olenka [21]2 years ago
5 0

Complete the square in the denominator.

s^2 - 4s + 8 = (s^2 - 4s + 4) + 4 = (s-2)^2 + 4

Rewrite the given transform as

\dfrac{3s-14}{s^2-4s+8} = \dfrac{3(s-2) - 8}{(s-2)^2+4} = 3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}

Now take the inverse transform:

L^{-1}_t\left\{3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3L^{-1}_t\left\{\dfrac{s-2}{(s-2)^2+2^2}\right\} - 4L^{-1}_t\left\{\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3e^{2t} L^{-1}_t\left\{\dfrac s{s^2+2^2}\right\} - 4e^{2t} L^{-1}_t\left\{\dfrac{2}{s^2+2^2}\right\} \\\\ \boxed{3e^{2t} \cos(2t) - 4e^{2t} \sin(2t)}

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<h2>Question:</h2>

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