What is inverse oflaplace 8/(3s^2+12)+3/(s^2-49) ?

The answer to inverse oflaplace 8/(3s^2+12)+3/(s^2-49) is 4/3 sin(2t)-3/14 e^{-7t}+3/14 e^{7t}

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inverse oflaplace 8/(3s^2+12)+3/(s^2-49)

inverse laplace

\frac{8}{3 s ^{2} + 12} + \frac{3}{s ^{2} - 49}

\frac{4}{3} sin (2 t) - \frac{3}{14} e^{- 7 t} + \frac{3}{14} e^{7 t}

Solution steps

L^{- 1} {\frac{8}{3 s ^{2} + 12} + \frac{3}{s ^{2} - 49}}

Take the partial fraction of

\frac{3}{s ^{2} - 49} : - \frac{3}{14 ( s + 7 )} + \frac{3}{14 ( s - 7 )}

= L^{- 1} {\frac{8}{3 s ^{2} + 12} - \frac{3}{14 ( s + 7 )} + \frac{3}{14 ( s - 7 )}}

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{8}{3 s ^{2} + 12}} - L^{- 1} {\frac{3}{14 ( s + 7 )}} + L^{- 1} {\frac{3}{14 ( s - 7 )}}

L^{- 1} {\frac{8}{3 s ^{2} + 12}} : \frac{4}{3} sin (2 t)

L^{- 1} {\frac{3}{14 ( s + 7 )}} : \frac{3}{14} e^{- 7 t}

L^{- 1} {\frac{3}{14 ( s - 7 )}} : \frac{3}{14} e^{7 t}

= \frac{4}{3} sin (2 t) - \frac{3}{14} e^{- 7 t} + \frac{3}{14} e^{7 t}

\int 1 + 64 x^{2} d x

\int e^{2} cos (x) d x

\frac{4}{s ^{4} + 4}

\int \frac{1}{u + 1} d u

\frac{d}{d x} (sin (9 x) + sin^{2} (9 x))

What is inverse oflaplace 8/(3s^2+12)+3/(s^2-49) ?
The answer to inverse oflaplace 8/(3s^2+12)+3/(s^2-49) is 4/3 sin(2t)-3/14 e^{-7t}+3/14 e^{7t}