What is inverse oflaplace (4s+7)/(s^2-4) ?

The answer to inverse oflaplace (4s+7)/(s^2-4) is 1/4 e^{-2t}+15/4 e^{2t}

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inverse oflaplace (4s+7)/(s^2-4)

inverse laplace

\frac{4 s + 7}{s ^{2} - 4}

\frac{1}{4} e^{- 2 t} + \frac{15}{4} e^{2 t}

Solution steps

L^{- 1} {\frac{4 s + 7}{s ^{2} - 4}}

Take the partial fraction of

\frac{4 s + 7}{s ^{2} - 4} : \frac{1}{4 ( s + 2 )} + \frac{15}{4 ( s - 2 )}

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

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

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

= \frac{1}{4} e^{- 2 t} + \frac{15}{4} e^{2 t}

\int \frac{x ^{2}}{x ^{2} + 1} d x

(y + 1) (\frac{d y}{d x}) = 2 x

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = - \frac{1}{2}, y^{^{'}} (0) = - \frac{9}{4}

\int \frac{15 e ^{5 y}}{e ^{5 y} + 2} d y

x^{3} + y^{3} = 8 x y, (4, 4)

What is inverse oflaplace (4s+7)/(s^2-4) ?
The answer to inverse oflaplace (4s+7)/(s^2-4) is 1/4 e^{-2t}+15/4 e^{2t}