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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

\frac{\partial}{\partial x} (e^{2 x + y})

y = f (x) = tan (x + 1)

f (x) = 7 x - 6 x, at x = 1

x \to 10 lim (12 x)

3 x^{2} y d x + (y + x^{3}) d y = 0

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