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

The answer to inverse oflaplace (2a+b)/((s+1)) is 2a+be^{-t}

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inverse oflaplace (2a+b)/((s+1))

inverse laplace

\frac{2 a + b}{( s + 1 )}

2 a + b e^{- t}

Solution steps

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

= L^{- 1} {2 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)}

= 2 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}

= 2 a + b e^{- t}

\int 20 + 75 e^{- \frac{x}{50}} d x

5 x^{2}

n = 1 \sum \infty (- 3)^{3 n - 1}

\frac{d}{d x} (ln (\frac{x ^{3} + 1}{x ^{2} + 1}))

\int 3 e^{3 t} d t

What is inverse oflaplace (2a+b)/((s+1)) ?
The answer to inverse oflaplace (2a+b)/((s+1)) is 2a+be^{-t}