What is inverse oflaplace (2s+6)/(8s^2-2s-3) ?

The answer to inverse oflaplace (2s+6)/(8s^2-2s-3) is -1/2 e^{-t/2}+3/4 e^{(3t)/4}

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inverse oflaplace (2s+6)/(8s^2-2s-3)

inverse laplace

\frac{2 s + 6}{8 s ^{2} - 2 s - 3}

- \frac{1}{2} e^{- \frac{t}{2}} + \frac{3}{4} e^{\frac{3 t}{4}}

Solution steps

L^{- 1} {\frac{2 s + 6}{8 s ^{2} - 2 s - 3}}

Take the partial fraction of

\frac{2 s + 6}{8 s ^{2} - 2 s - 3} : - \frac{1}{2 s + 1} + \frac{3}{4 s - 3}

= L^{- 1} {- \frac{1}{2 s + 1} + \frac{3}{4 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}{2 s + 1}} + L^{- 1} {\frac{3}{4 s - 3}}

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

L^{- 1} {\frac{3}{4 s - 3}} : \frac{3}{4} e^{\frac{3 t}{4}}

= - \frac{1}{2} e^{- \frac{t}{2}} + \frac{3}{4} e^{\frac{3 t}{4}}

y = 2 x^{2}, y^{2} = \frac{1}{2} x

y^{^{'}} y^{- 2} + x y^{- 1} = x

\frac{5 x + 2}{12 x + 5}

\int \frac{9 x ^{2} - 1}{x ^{2}} d x

\frac{d}{d x} (\frac{100}{x ^{2}})

What is inverse oflaplace (2s+6)/(8s^2-2s-3) ?
The answer to inverse oflaplace (2s+6)/(8s^2-2s-3) is -1/2 e^{-t/2}+3/4 e^{(3t)/4}