What is inverse oflaplace 5/((s+1)(s+2)^2) ?

The answer to inverse oflaplace 5/((s+1)(s+2)^2) is 5e^{-t}-5e^{-2t}-e^{-2t}*5t

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inverse oflaplace 5/((s+1)(s+2)^2)

inverse laplace

\frac{5}{( s + 1 ) ( s + 2 ) ^{2}}

5 e^{- t} - 5 e^{- 2 t} - e^{- 2 t} \cdot 5 t

Solution steps

L^{- 1} {\frac{5}{( s + 1 ) ( s + 2 ) ^{2}}}

Take the partial fraction of

\frac{5}{( s + 1 ) ( s + 2 ) ^{2}} : \frac{5}{s + 1} - \frac{5}{s + 2} - \frac{5}{( s + 2 ) ^{2}}

= L^{- 1} {\frac{5}{s + 1} - \frac{5}{s + 2} - \frac{5}{( s + 2 ) ^{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{5}{s + 1}} - L^{- 1} {\frac{5}{s + 2}} - L^{- 1} {\frac{5}{( s + 2 ) ^{2}}}

L^{- 1} {\frac{5}{s + 1}} : 5 e^{- t}

L^{- 1} {\frac{5}{s + 2}} : 5 e^{- 2 t}

L^{- 1} {\frac{5}{( s + 2 ) ^{2}}} : e^{- 2 t} \cdot 5 t

= 5 e^{- t} - 5 e^{- 2 t} - e^{- 2 t} \cdot 5 t

cos (3 \cdot x) \cdot cos (5 \cdot x)

\frac{\partial}{\partial y} (e^{(\frac{x}{y})})

\frac{d}{d x} (7^{9 x})

\int x^{3} \cdot x^{2} + 1 d x

x^{3} - 8 x^{2} + 5 x + 8

What is inverse oflaplace 5/((s+1)(s+2)^2) ?
The answer to inverse oflaplace 5/((s+1)(s+2)^2) is 5e^{-t}-5e^{-2t}-e^{-2t}*5t