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

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

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inverse oflaplace 3/(s^2-49)

inverse laplace

\frac{3}{s ^{2} - 49}

- \frac{3}{14} e^{- 7 t} + \frac{3}{14} e^{7 t}

Solution steps

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

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{3}{14} e^{- 7 t} + \frac{3}{14} e^{7 t}

\frac{d ^{2}}{d x ^{2}} (x 4 - x^{2})

\frac{d y}{d x}, y = 11 x

2 x^{2} y + x^{3} y^{^{'}} = 1

\int \frac{16}{64 + ( 2 π x ) ^{2}} d x

y^{^{'}} (x^{2} + 1) + 3 x y = 6 x

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