What is inverse oflaplace 3/((s+1)) ?

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

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inverse oflaplace 3/((s+1))

inverse laplace

\frac{3}{( s + 1 )}

3 e^{- t}

Solution steps

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

= L^{- 1} {3 \cdot \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)}

= 3 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}

= 3 e^{- t}

x \to - 1 lim (5^{x^{2} - 3 x})

\int \frac{x ^{3}}{( x ^{4} + 3 ) ^{\frac{7}{2}}} d x

\int \frac{e ^{x}}{e ^{2 x} + 4} d x

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

(e^{2 x} - 5 x)^{2}

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