What is inverse oflaplace ((3s+5))/((s-5)(s+2)) ?

The answer to inverse oflaplace ((3s+5))/((s-5)(s+2)) is 20/7 e^{5t}+1/7 e^{-2t}

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inverse oflaplace ((3s+5))/((s-5)(s+2))

inverse laplace

\frac{( 3 s + 5 )}{( s - 5 ) ( s + 2 )}

\frac{20}{7} e^{5 t} + \frac{1}{7} e^{- 2 t}

Solution steps

L^{- 1} {\frac{( 3 s + 5 )}{( s - 5 ) ( s + 2 )}}

Take the partial fraction of

\frac{3 s + 5}{( s - 5 ) ( s + 2 )} : \frac{20}{7 ( s - 5 )} + \frac{1}{7 ( s + 2 )}

= L^{- 1} {\frac{20}{7 ( s - 5 )} + \frac{1}{7 ( s + 2 )}}

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{20}{7 ( s - 5 )}} + L^{- 1} {\frac{1}{7 ( s + 2 )}}

L^{- 1} {\frac{20}{7 ( s - 5 )}} : \frac{20}{7} e^{5 t}

L^{- 1} {\frac{1}{7 ( s + 2 )}} : \frac{1}{7} e^{- 2 t}

= \frac{20}{7} e^{5 t} + \frac{1}{7} e^{- 2 t}

\int \frac{4 x}{x ^{2} + 9} d x

\frac{d}{d x} ((x^{3} + 1) e^{x})

\frac{1}{2 x + 2}

\frac{d}{d x} (\frac{ln ( 7 x )}{7 x})

\int_{0}^{\infty} e^{- x} x^{2} d x

What is inverse oflaplace ((3s+5))/((s-5)(s+2)) ?
The answer to inverse oflaplace ((3s+5))/((s-5)(s+2)) is 20/7 e^{5t}+1/7 e^{-2t}