What is inverse oflaplace (-a-b)/((s+1)) ?

The answer to inverse oflaplace (-a-b)/((s+1)) is -a-be^{-t}

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inverse oflaplace (-a-b)/((s+1))

inverse laplace

\frac{- a - b}{( s + 1 )}

- a - b e^{- t}

Solution steps

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

= L^{- 1} {- a - b \frac{1}{s + 1}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= - a - b L^{- 1} {\frac{1}{s + 1}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

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

= - a - b e^{- t}

\frac{2}{3} x^{\frac{3}{2}}

\frac{3}{s ( s + 2 )}

\int \frac{1}{8 x} d x

y^{^{''}} - 5 y^{^{'}} + 6 y = 36 t

x \frac{d}{d x} [ln (y)] = 1

What is inverse oflaplace (-a-b)/((s+1)) ?
The answer to inverse oflaplace (-a-b)/((s+1)) is -a-be^{-t}