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

The answer to inverse oflaplace (9s)/((s+9)^2) is 9e^{-9t}-e^{-9t}*81t

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

inverse laplace

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

9 e^{- 9 t} - e^{- 9 t} \cdot 81 t

Solution steps

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

Take the partial fraction of

\frac{9 s}{( s + 9 ) ^{2}} : \frac{9}{s + 9} - \frac{81}{( s + 9 ) ^{2}}

= L^{- 1} {\frac{9}{s + 9} - \frac{81}{( s + 9 ) ^{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{9}{s + 9}} - L^{- 1} {\frac{81}{( s + 9 ) ^{2}}}

L^{- 1} {\frac{9}{s + 9}} : 9 e^{- 9 t}

L^{- 1} {\frac{81}{( s + 9 ) ^{2}}} : e^{- 9 t} \cdot 81 t

= 9 e^{- 9 t} - e^{- 9 t} \cdot 81 t

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What is inverse oflaplace (9s)/((s+9)^2) ?
The answer to inverse oflaplace (9s)/((s+9)^2) is 9e^{-9t}-e^{-9t}*81t