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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

- \frac{5}{2} x^{2} + 2 x + 2

\int 3 x^{2} y^{2} d x

\int_{3}^{5} (x^{4} - \frac{3}{x ^{3}}) d x

2 (ln (y)) y^{^{'}} - x^{5} y = 0

y = cos (a^{5} + x^{5})

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