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

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

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

inverse laplace

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

H (t) - e^{- t} + e^{- t} t

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{s}} : H (t)

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

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

= H (t) - e^{- t} + e^{- t} t

y = x^{s i n (x)}

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = - 4, y^{^{'}} (0) = 0

6 x^{2} - x^{3}

\int \frac{5.45}{( 1 + 1.9 x ) ^{0.9}} d x

f (x) = ln (4 - x^{6})

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