Question

The Laplace transform of t³ is​

Answer 1

Answer:

cos(4t) = f(t), say. Now if L(f(t)) = F(s), then, by theorem on derivatives of Laplace transforms, we have L(t*f(t)) = -F'(s), where the ' denotes derivative with respect to s. We know that

L(cos(4t)) = s/(s^2+16). Therefore in view of the above theorem, we get that L(t*cos(4t)) = -[s/(s^2 +16)]' =

-[(s^2 +16)*1 -s.2s]/(s^2 +16)^2 =

(s^2–16)/(s^2 +16)^2. Similarly applying the theorem twice more we get L(t^3.cos(4t)).

Step-by-step explanation:

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