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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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