What is inverse oflaplace (3s)/((s+3)^2(s+6)) ?

The answer to inverse oflaplace (3s)/((s+3)^2(s+6)) is 2e^{-3t}-e^{-3t}*3t-2e^{-6t}

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inverse oflaplace (3s)/((s+3)^2(s+6))

inverse laplace

\frac{3 s}{( s + 3 ) ^{2} ( s + 6 )}

2 e^{- 3 t} - e^{- 3 t} \cdot 3 t - 2 e^{- 6 t}

Solution steps

L^{- 1} {\frac{3 s}{( s + 3 ) ^{2} ( s + 6 )}}

Take the partial fraction of

\frac{3 s}{( s + 3 ) ^{2} ( s + 6 )} : \frac{2}{s + 3} - \frac{3}{( s + 3 ) ^{2}} - \frac{2}{s + 6}

= L^{- 1} {\frac{2}{s + 3} - \frac{3}{( s + 3 ) ^{2}} - \frac{2}{s + 6}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{2}{s + 3}} - L^{- 1} {\frac{3}{( s + 3 ) ^{2}}} - L^{- 1} {\frac{2}{s + 6}}

L^{- 1} {\frac{2}{s + 3}} : 2 e^{- 3 t}

L^{- 1} {\frac{3}{( s + 3 ) ^{2}}} : e^{- 3 t} \cdot 3 t

L^{- 1} {\frac{2}{s + 6}} : 2 e^{- 6 t}

= 2 e^{- 3 t} - e^{- 3 t} \cdot 3 t - 2 e^{- 6 t}

\int_{0}^{l n (5)} 4 (5 - x e^{x}) d x

x \to - \infty lim (\frac{4}{x})

\int (\frac{sin ( x )}{x}) d x

f (x) = x^{7} + x^{5} + x^{3} + x + 1

(3.3) (4.5)

What is inverse oflaplace (3s)/((s+3)^2(s+6)) ?
The answer to inverse oflaplace (3s)/((s+3)^2(s+6)) is 2e^{-3t}-e^{-3t}*3t-2e^{-6t}