What is inverse oflaplace 9/(s(s+3)^2) ?

The answer to inverse oflaplace 9/(s(s+3)^2) is \H(t)-e^{-3t}-e^{-3t}*3t

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inverse oflaplace 9/(s(s+3)^2)

inverse laplace

\frac{9}{s ( s + 3 ) ^{2}}

H (t) - e^{- 3 t} - e^{- 3 t} \cdot 3 t

Solution steps

L^{- 1} {\frac{9}{s ( s + 3 ) ^{2}}}

Take the partial fraction of

\frac{9}{s ( s + 3 ) ^{2}} : \frac{1}{s} - \frac{1}{s + 3} - \frac{3}{( s + 3 ) ^{2}}

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

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

L^{- 1} {\frac{1}{s}} : H (t)

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

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

= H (t) - e^{- 3 t} - e^{- 3 t} \cdot 3 t

\frac{d}{d x} (- 3 sin (sin (3 x)) cos (3 x))

\int_{0}^{2} (x^{2} + 1) d x

\frac{d}{d x} (x^{5} - 3 x^{4} + \frac{4}{3 x ^{2}} + \frac{5}{x ^{3}})

\frac{d}{d x} (2 \cdot cos (2 x))

\frac{\partial}{\partial x} (cos (π x) - x^{2} y + e^{x z} + yz)

What is inverse oflaplace 9/(s(s+3)^2) ?
The answer to inverse oflaplace 9/(s(s+3)^2) is \H(t)-e^{-3t}-e^{-3t}*3t