What is inverse oflaplace 1/(3s+4) ?

The answer to inverse oflaplace 1/(3s+4) is 1/3 e^{-(4t)/3}

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inverse oflaplace 1/(3s+4)

inverse laplace

\frac{1}{3 s + 4}

\frac{1}{3} e^{- \frac{4 t}{3}}

Solution steps

L^{- 1} {\frac{1}{3 s + 4}}

= L^{- 1} {\frac{1}{3} \cdot \frac{1}{s + \frac{4}{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{1}{3} L^{- 1} {\frac{1}{s + \frac{4}{3}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + \frac{4}{3}}} = e^{- \frac{4 t}{3}}

= \frac{1}{3} e^{- \frac{4 t}{3}}

\int \frac{4}{x ^{2} x ^{2} + 36} d x

y^{^{''}} - 8 y^{^{'}} + 19 y = 0, y (0) = - 4, y^{^{'}} (0) = 2

\int_{1}^{2} \frac{x}{x ^{2} + 1} d x

(x - \frac{2}{x})^{3}

\int_{0}^{7} e^{x} - 7 d x

What is inverse oflaplace 1/(3s+4) ?
The answer to inverse oflaplace 1/(3s+4) is 1/3 e^{-(4t)/3}