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

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

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inverse oflaplace 4/(s+3)

inverse laplace

\frac{4}{s + 3}

4 e^{- 3 t}

Solution steps

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

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

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

Use Inverse Laplace Transform table:

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

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

= 4 e^{- 3 t}

x \to \infty lim (- x e^{- x^{2}} 2 - e^{- x^{2}})

\frac{d v}{d y} + 2 \cdot v = 4 e^{- y}

\int_{- 8}^{3} ∣ x ∣ + 1 d x

g (t) = t

\int \frac{ln ( 4 x )}{x ^{2}} d x

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