What is inverse oflaplace 1/((s^2+1)(s^2+16)) ?

The answer to inverse oflaplace 1/((s^2+1)(s^2+16)) is 1/15 sin(t)-1/60 sin(4t)

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inverse oflaplace 1/((s^2+1)(s^2+16))

inverse laplace

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

\frac{1}{15} sin (t) - \frac{1}{60} sin (4 t)

Solution steps

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

Take the partial fraction of

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 16 )} : \frac{1}{15 ( s ^{2} + 1 )} - \frac{1}{15 ( s ^{2} + 16 )}

= L^{- 1} {\frac{1}{15 ( s ^{2} + 1 )} - \frac{1}{15 ( 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{1}{15 ( s ^{2} + 1 )}} - L^{- 1} {\frac{1}{15 ( s ^{2} + 16 )}}

L^{- 1} {\frac{1}{15 ( s ^{2} + 1 )}} : \frac{1}{15} sin (t)

L^{- 1} {\frac{1}{15 ( s ^{2} + 16 )}} : \frac{1}{60} sin (4 t)

= \frac{1}{15} sin (t) - \frac{1}{60} sin (4 t)

x \to 2 lim (x^{2} - 4 x + 5)

\frac{d}{d x} (- \frac{2 x}{( x ^{2} - 1 ) ^{2}})

\int 21 x^{3} e^{x} d x

\frac{\partial}{\partial x} (3 z^{3} cot (x z))

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

What is inverse oflaplace 1/((s^2+1)(s^2+16)) ?
The answer to inverse oflaplace 1/((s^2+1)(s^2+16)) is 1/15 sin(t)-1/60 sin(4t)