What is inverse oflaplace 2/((3s+1)) ?

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

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inverse oflaplace 2/((3s+1))

inverse laplace

\frac{2}{( 3 s + 1 )}

\frac{2}{3} e^{- \frac{t}{3}}

Solution steps

L^{- 1} {\frac{2}{( 3 s + 1 )}}

= L^{- 1} {\frac{2}{3} \cdot \frac{1}{s + \frac{1}{3}}}

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)}

= \frac{2}{3} L^{- 1} {\frac{1}{s + \frac{1}{3}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + \frac{1}{3}}} = e^{- \frac{t}{3}}

= \frac{2}{3} e^{- \frac{t}{3}}

\int_{- 2}^{2} \frac{e ^{x}}{e ^{x} - 1} d x

\frac{d}{d x} (x e^{x s i n (x)})

\int \frac{sin ( u )}{u} d u

\int \frac{x - 15}{x ^{2} ( x + 3 )} d x

2 y^{^{''}} - 16 y^{^{'}} + 32 y = 4 e^{4 t}

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