What is inverse oflaplace s/((s^2+25)(s^2+16)) ?

The answer to inverse oflaplace s/((s^2+25)(s^2+16)) is -1/9 cos(5t)+1/9 cos(4t)

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inverse oflaplace s/((s^2+25)(s^2+16))

inverse laplace

\frac{s}{( s ^{2} + 25 ) ( s ^{2} + 16 )}

- \frac{1}{9} cos (5 t) + \frac{1}{9} cos (4 t)

Solution steps

L^{- 1} {\frac{s}{( s ^{2} + 25 ) ( s ^{2} + 16 )}}

Take the partial fraction of

\frac{s}{( s ^{2} + 25 ) ( s ^{2} + 16 )} : - \frac{s}{9 ( s ^{2} + 25 )} + \frac{s}{9 ( s ^{2} + 16 )}

= L^{- 1} {- \frac{s}{9 ( s ^{2} + 25 )} + \frac{s}{9 ( 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}{9 ( s ^{2} + 25 )}} + L^{- 1} {\frac{s}{9 ( s ^{2} + 16 )}}

L^{- 1} {\frac{s}{9 ( s ^{2} + 25 )}} : \frac{1}{9} cos (5 t)

L^{- 1} {\frac{s}{9 ( s ^{2} + 16 )}} : \frac{1}{9} cos (4 t)

= - \frac{1}{9} cos (5 t) + \frac{1}{9} cos (4 t)

\int x^{2} (4 x^{3} - 3)^{4} d x

\int - 6 sec^{2} (x) d x

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\int 5 e^{x} sin^{2} (x) d x

\int 9 cot^{3} (x) csc^{3} (x) d x

What is inverse oflaplace s/((s^2+25)(s^2+16)) ?
The answer to inverse oflaplace s/((s^2+25)(s^2+16)) is -1/9 cos(5t)+1/9 cos(4t)