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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

x y y^{^{'}} = y^{2} + x 4 x^{2} + y^{2}

e^{w}

x^{3} - 2 x^{2} - 5 x + 6, at x = 2

\int 8 x^{- 9} d x

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