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

The answer to inverse oflaplace (-3s+9)/((s+3)^2) is -3e^{-3t}+e^{-3t}*18t

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

inverse laplace

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

- 3 e^{- 3 t} + e^{- 3 t} \cdot 18 t

Solution steps

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

Take the partial fraction of

\frac{- 3 s + 9}{( s + 3 ) ^{2}} : - \frac{3}{s + 3} + \frac{18}{( s + 3 ) ^{2}}

= L^{- 1} {- \frac{3}{s + 3} + \frac{18}{( 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{3}{s + 3}} + L^{- 1} {\frac{18}{( s + 3 ) ^{2}}}

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

L^{- 1} {\frac{18}{( s + 3 ) ^{2}}} : e^{- 3 t} \cdot 18 t

= - 3 e^{- 3 t} + e^{- 3 t} \cdot 18 t

\frac{d}{d t} (4 e^{t} sin (t))

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

\int_{5}^{8} \frac{6}{x - 5} d x

t^{2} y^{^{'}} + y^{2} = t y, y (1) = - \frac{1}{2}

\frac{\partial}{\partial x} (ln (x - y^{2}))

What is inverse oflaplace (-3s+9)/((s+3)^2) ?
The answer to inverse oflaplace (-3s+9)/((s+3)^2) is -3e^{-3t}+e^{-3t}*18t