What is inverse oflaplace s/((s^2+16)(s^2+9)) ?

The answer to inverse oflaplace s/((s^2+16)(s^2+9)) is -1/7 cos(4t)+1/7 cos(3t)

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inverse oflaplace s/((s^2+16)(s^2+9))

inverse laplace

\frac{s}{( s ^{2} + 16 ) ( s ^{2} + 9 )}

- \frac{1}{7} cos (4 t) + \frac{1}{7} cos (3 t)

Solution steps

L^{- 1} {\frac{s}{( s ^{2} + 16 ) ( s ^{2} + 9 )}}

Take the partial fraction of

\frac{s}{( s ^{2} + 16 ) ( s ^{2} + 9 )} : - \frac{s}{7 ( s ^{2} + 16 )} + \frac{s}{7 ( s ^{2} + 9 )}

= L^{- 1} {- \frac{s}{7 ( s ^{2} + 16 )} + \frac{s}{7 ( s ^{2} + 9 )}}

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}{7 ( s ^{2} + 16 )}} + L^{- 1} {\frac{s}{7 ( s ^{2} + 9 )}}

L^{- 1} {\frac{s}{7 ( s ^{2} + 16 )}} : \frac{1}{7} cos (4 t)

L^{- 1} {\frac{s}{7 ( s ^{2} + 9 )}} : \frac{1}{7} cos (3 t)

= - \frac{1}{7} cos (4 t) + \frac{1}{7} cos (3 t)

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 3, y^{^{'}} (0) = 4

\frac{d}{d x} (x - ln (8 x))

(7 x + 4)^{2}

\frac{d y}{d x} = 3 x y

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

What is inverse oflaplace s/((s^2+16)(s^2+9)) ?
The answer to inverse oflaplace s/((s^2+16)(s^2+9)) is -1/7 cos(4t)+1/7 cos(3t)