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

The answer to inverse oflaplace (s-1)/(s(s+4)+3) is -e^{-t}+2e^{-3t}

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inverse oflaplace (s-1)/(s(s+4)+3)

inverse laplace

\frac{s - 1}{s ( s + 4 ) + 3}

- e^{- t} + 2 e^{- 3 t}

Solution steps

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

Take the partial fraction of

\frac{s - 1}{s ( s + 4 ) + 3} : - \frac{1}{s + 1} + \frac{2}{s + 3}

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

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

L^{- 1} {\frac{2}{s + 3}} : 2 e^{- 3 t}

= - e^{- t} + 2 e^{- 3 t}

y^{^{''}} + 2 y = 0

x \to \infty lim (10 + 3 x)

\frac{d}{d x} ((x - \frac{1}{x ^{2}})^{5})

\int \frac{11 + u}{u} d u

\int_{0}^{9} 81 + x^{2} d x

What is inverse oflaplace (s-1)/(s(s+4)+3) ?
The answer to inverse oflaplace (s-1)/(s(s+4)+3) is -e^{-t}+2e^{-3t}