What is inverse oflaplace s/((s^2+49)(s^2+1)) ?

The answer to inverse oflaplace s/((s^2+49)(s^2+1)) is -1/48 cos(7t)+1/48 cos(t)

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inverse oflaplace s/((s^2+49)(s^2+1))

inverse laplace

\frac{s}{( s ^{2} + 49 ) ( s ^{2} + 1 )}

- \frac{1}{48} cos (7 t) + \frac{1}{48} cos (t)

Solution steps

L^{- 1} {\frac{s}{( s ^{2} + 49 ) ( s ^{2} + 1 )}}

Take the partial fraction of

\frac{s}{( s ^{2} + 49 ) ( s ^{2} + 1 )} : - \frac{s}{48 ( s ^{2} + 49 )} + \frac{s}{48 ( s ^{2} + 1 )}

= L^{- 1} {- \frac{s}{48 ( s ^{2} + 49 )} + \frac{s}{48 ( s ^{2} + 1 )}}

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{s}{48 ( s ^{2} + 49 )}} + L^{- 1} {\frac{s}{48 ( s ^{2} + 1 )}}

L^{- 1} {\frac{s}{48 ( s ^{2} + 49 )}} : \frac{1}{48} cos (7 t)

L^{- 1} {\frac{s}{48 ( s ^{2} + 1 )}} : \frac{1}{48} cos (t)

= - \frac{1}{48} cos (7 t) + \frac{1}{48} cos (t)

\frac{x}{x ^{2} - 9}

\frac{d}{d y} (sinh (y))

x^{2} + 11 x y + y^{2} = 11

x (x^{3} + 1) y^{^{'}} + (2 x^{3} - 1) y = \frac{( x ^{3} - 2 )}{x}

\frac{d}{d x} (3 lo g_{10} (x))

What is inverse oflaplace s/((s^2+49)(s^2+1)) ?
The answer to inverse oflaplace s/((s^2+49)(s^2+1)) is -1/48 cos(7t)+1/48 cos(t)