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

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

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inverse oflaplace 2/((s+3))

inverse laplace

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

2 e^{- 3 t}

Solution steps

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

= L^{- 1} {2 \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)}

= 2 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}

= 2 e^{- 3 t}

\frac{d}{d x} (\frac{1}{sin ( 3 x )})

\frac{d}{d x} (\frac{x ^{2} + 2 x - 6}{( x + 1 ) ^{2}})

(x^{2} + 1) y^{^{''}} - 2 x y^{^{'}} + 2 y = 0

f (x) = 6 e^{x} + x, (0, 6)

\frac{d}{d x} (cos^{4} (2 x^{3} - 5))

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