What is inverse oflaplace ((s+1))/((s+6)) ?

The answer to inverse oflaplace ((s+1))/((s+6)) is \delta(t)-5e^{-6t}

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inverse oflaplace ((s+1))/((s+6))

inverse laplace

\frac{( s + 1 )}{( s + 6 )}

δ (t) - 5 e^{- 6 t}

Solution steps

L^{- 1} {\frac{( s + 1 )}{( s + 6 )}}

Take the partial fraction of

\frac{s + 1}{s + 6} : 1 - \frac{5}{s + 6}

= L^{- 1} {1 - \frac{5}{s + 6}}

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} {1} - L^{- 1} {\frac{5}{s + 6}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{5}{s + 6}} : 5 e^{- 6 t}

= δ (t) - 5 e^{- 6 t}

\int \frac{x}{x + y} d x

\frac{d y}{d x}, y^{2} = 12 x

\int 2 x e - 2 x d x

\int x^{2} + y d x

y = (x - 1)^{3} (x + 2)^{4}

What is inverse oflaplace ((s+1))/((s+6)) ?
The answer to inverse oflaplace ((s+1))/((s+6)) is \delta(t)-5e^{-6t}