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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

f (x) = 5 x^{9} x^{6}, at x = 2

y^{^{'}} = \frac{y ^{4} - 3}{4 x y ^{3}}

f (x) = x^{\frac{3}{5}}

\frac{x ^{2} - a ^{2}}{x - a}

y = x^{6 x}

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