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Find the inverse laplace transform of: (2 s + 4) / (s - 3)^3

User Maclir
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1 Answer

6 votes

Answer:


e^(3t)(2t+5t^(2))

Explanation:


L^(-1)[(2s+4)/((s-3)^(3)) ]=

Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3

Translation theorem:
L^(1) [F(s-a)=L^(-1)[F(s)|_(s \to s-a)\\ L^(1) [F(s-a)=e^(at) f(t)


L^(-1)[(2s+4)/((s-3)^(3)) ]=e^(3t) L^(-1)[(2(s+3)+4)/(s^(3)) ]

Separate the fraction in a sum:


e^(3t) L^(-1)[(2s+10)/(s^(3)) ]=e^(3t) L^(-1)[(2s)/(s^(3))+(10)/(s^(3)) ]=e^(3t) (L^(-1)[(2)/(s^(2))]+ L^(-1)[(10)/(s^(3))])

The formula for this is:


L^(-1)[(n!)/(s^(n+1)) ]=t^(n)

Modify the expression to match the formula.


e^(3t) (2L^(-1)[(1)/(s^(1+1))]+ (10)/(2) L^(-1)[(2)/(s^(2+1))])=e^(3t) (2L^(-1)[(1)/(s^(1+1))]+ 5 L^(-1)[(2)/(s^(2+1))])

Solve


e^(3t) (2L^(-1)[(1)/(s^(1+1))]+ 5 L^(-1)[(2)/(s^(2+1))])=e^(3t)(2t+5t^(2) )

User Chui Tey
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