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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

ln (\frac{1 + x}{1 - x})

\frac{\partial}{\partial y} (4 x^{3} y^{3} - 2)

f (x) = \frac{4 + cot ( x )}{7 - tan ( x )}

5 t

\frac{6}{s ( s + 1 ) ( s + 2 ) + 6}

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