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

The answer to inverse oflaplace (s+1)/(s^2+2s+s) is Undefined+1/3 \H(t)-1/3 e^{-3t}

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

inverse laplace

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

Undefined + \frac{1}{3} H (t) - \frac{1}{3} e^{- 3 t}

Solution steps

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

Factor

s^{2} + 2 s + s : s (s + 3)

= L^{- 1} {\frac{s + 1}{s ( s + 3 )}}

Expand

= L^{- 1} {\frac{s}{s ( s + 3 )} + \frac{1}{s ( s + 3 )}}

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{s}{s ( s + 3 )}} + L^{- 1} {\frac{1}{s ( s + 3 )}}

L^{- 1} {\frac{s}{s ( s + 3 )}} :

Undefined

L^{- 1} {\frac{1}{s ( s + 3 )}} : \frac{1}{3} H (t) - \frac{1}{3} e^{- 3 t}

= Undefined + \frac{1}{3} H (t) - \frac{1}{3} e^{- 3 t}

\frac{\partial}{\partial x} (\frac{6 x}{y})

\int sin (π y) d y

\frac{\partial}{\partial x} (sin (x^{2} - y^{2}))

11 x^{2} - 4 x

y^{^{'}} = \frac{( sec ( y ) )}{( x - 4 ) ^{2}}

What is inverse oflaplace (s+1)/(s^2+2s+s) ?
The answer to inverse oflaplace (s+1)/(s^2+2s+s) is Undefined+1/3 \H(t)-1/3 e^{-3t}