What is inverse oflaplace (5x+5)/(x^2-4) ?

The answer to inverse oflaplace (5x+5)/(x^2-4) is 5/4 e^{-2t}+15/4 e^{2t}

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inverse oflaplace (5x+5)/(x^2-4)

inverse laplace

\frac{5 x + 5}{x ^{2} - 4}

\frac{5}{4} e^{- 2 t} + \frac{15}{4} e^{2 t}

Solution steps

L^{- 1} {\frac{5 x + 5}{x ^{2} - 4}}

Take the partial fraction of

\frac{5 x + 5}{x ^{2} - 4} : \frac{5}{4 ( x + 2 )} + \frac{15}{4 ( x - 2 )}

= L^{- 1} {\frac{5}{4 ( x + 2 )} + \frac{15}{4 ( x - 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{5}{4 ( x + 2 )}} + L^{- 1} {\frac{15}{4 ( x - 2 )}}

L^{- 1} {\frac{5}{4 ( x + 2 )}} : \frac{5}{4} e^{- 2 t}

L^{- 1} {\frac{15}{4 ( x - 2 )}} : \frac{15}{4} e^{2 t}

= \frac{5}{4} e^{- 2 t} + \frac{15}{4} e^{2 t}

\int \frac{e ^{l n (x)}}{x ^{2}} d x

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

x \to 4 lim (x 20 - x^{2})

\frac{\partial}{\partial x} (12 x y^{2})

\int_{0}^{1} ((x^{2} + 1) e^{- x}) d x

What is inverse oflaplace (5x+5)/(x^2-4) ?
The answer to inverse oflaplace (5x+5)/(x^2-4) is 5/4 e^{-2t}+15/4 e^{2t}