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

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

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

inverse laplace

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

H (t) - e^{- 3 t}

Solution steps

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

Take the partial fraction of

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

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {\frac{1}{s}} : H (t)

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

= H (t) - e^{- 3 t}

x^{2} + 3 x

\frac{\partial}{\partial x} (x^{2} y^{3} + x y - 2)

\int_{l n (4)}^{l n (9)} e^{\frac{x}{2}} d x

f (x) = sin (\frac{1}{x ^{2}})

\int \frac{cos ^{6} ( x )}{sin ^{10} ( x )} d x

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