What is inverse oflaplace (3^2)/((s+3)^2)*1/s ?

The answer to inverse oflaplace (3^2)/((s+3)^2)*1/s is \H(t)-e^{-3t}-3e^{-3t}t

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inverse oflaplace (3^2)/((s+3)^2)*1/s

inverse laplace

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

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

Solution steps

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

Expand

\frac{3 ^{2}}{( s + 3 ) ^{2}} \frac{1}{s} : \frac{9}{s ^{3} + 6 s ^{2} + 9 s}

= L^{- 1} {\frac{9}{s ^{3} + 6 s ^{2} + 9 s}}

Take the partial fraction of

\frac{9}{s ^{3} + 6 s ^{2} + 9 s} : \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}}}

Expand

= L^{- 1} {\frac{1}{s} - \frac{1}{s + 3} - \frac{3}{s ^{2} + 6 s + 9}}

Expand

\frac{3}{s ^{2} + 6 s + 9} : 3 \cdot \frac{1}{( s + 3 ) ^{2}}

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

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

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

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

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

\int \frac{1 - w}{w} d w

f (x) = sin^{2} (4 x)

\frac{\partial}{\partial v} (2 sin (u) cos (v))

e^{x}, \frac{e ^{2 x}}{2}, 0, 1

\frac{d}{d x} (- \frac{1}{4 ( x ) ^{\frac{3}{2}}})

What is inverse oflaplace (3^2)/((s+3)^2)*1/s ?
The answer to inverse oflaplace (3^2)/((s+3)^2)*1/s is \H(t)-e^{-3t}-3e^{-3t}t