What is inverse oflaplace (s+2)/((s-1)(s+6)) ?

The answer to inverse oflaplace (s+2)/((s-1)(s+6)) is 3/7 e^t+4/7 e^{-6t}

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inverse oflaplace (s+2)/((s-1)(s+6))

inverse laplace

\frac{s + 2}{( s - 1 ) ( s + 6 )}

\frac{3}{7} e^{t} + \frac{4}{7} e^{- 6 t}

Solution steps

L^{- 1} {\frac{s + 2}{( s - 1 ) ( s + 6 )}}

Take the partial fraction of

\frac{s + 2}{( s - 1 ) ( s + 6 )} : \frac{3}{7 ( s - 1 )} + \frac{4}{7 ( s + 6 )}

= L^{- 1} {\frac{3}{7 ( s - 1 )} + \frac{4}{7 ( s + 6 )}}

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{3}{7 ( s - 1 )}} + L^{- 1} {\frac{4}{7 ( s + 6 )}}

L^{- 1} {\frac{3}{7 ( s - 1 )}} : \frac{3}{7} e^{t}

L^{- 1} {\frac{4}{7 ( s + 6 )}} : \frac{4}{7} e^{- 6 t}

= \frac{3}{7} e^{t} + \frac{4}{7} e^{- 6 t}

2 x, x^{2} - 4 x

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

\int_{0}^{1} \frac{2 x}{( x ^{2} + 2 ) ^{3}} d x

\frac{\partial}{\partial y} (tan (4 + 2 x^{2} y^{3} z^{2}))

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

What is inverse oflaplace (s+2)/((s-1)(s+6)) ?
The answer to inverse oflaplace (s+2)/((s-1)(s+6)) is 3/7 e^t+4/7 e^{-6t}