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

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

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

inverse laplace

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

e^{- t} - e^{- 2 t} - e^{- 2 t} t

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= e^{- t} - e^{- 2 t} - e^{- 2 t} t

\frac{1}{3 - 2 x}

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

(1 + e^{x}) y y^{^{'}} = e^{x}, y (0) = 1

f (x) = 3 x^{2} - 8 x - x

x \to - 3 lim (x^{3} - 4 x^{2} - 7)

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