What is inverse oflaplace 3/((s^2+9)(s^2+1)) ?

The answer to inverse oflaplace 3/((s^2+9)(s^2+1)) is -1/8 sin(3t)+3/8 sin(t)

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inverse oflaplace 3/((s^2+9)(s^2+1))

inverse laplace

\frac{3}{( s ^{2} + 9 ) ( s ^{2} + 1 )}

- \frac{1}{8} sin (3 t) + \frac{3}{8} sin (t)

Solution steps

L^{- 1} {\frac{3}{( s ^{2} + 9 ) ( s ^{2} + 1 )}}

Take the partial fraction of

\frac{3}{( s ^{2} + 9 ) ( s ^{2} + 1 )} : - \frac{3}{8 ( s ^{2} + 9 )} + \frac{3}{8 ( s ^{2} + 1 )}

= L^{- 1} {- \frac{3}{8 ( s ^{2} + 9 )} + \frac{3}{8 ( 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{3}{8 ( s ^{2} + 9 )}} + L^{- 1} {\frac{3}{8 ( s ^{2} + 1 )}}

L^{- 1} {\frac{3}{8 ( s ^{2} + 9 )}} : \frac{1}{8} sin (3 t)

L^{- 1} {\frac{3}{8 ( s ^{2} + 1 )}} : \frac{3}{8} sin (t)

= - \frac{1}{8} sin (3 t) + \frac{3}{8} sin (t)

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

2 y = 3 x, 2 y + 3 x = 6, y = 4

\int_{\frac{π}{2}}^{π} cos (x) d x

\int \frac{1}{e ^{x} + 4 e ^{- x}} d x

\int_{1}^{3} 2 π x (- x^{2} + 4 x - 3) d x

What is inverse oflaplace 3/((s^2+9)(s^2+1)) ?
The answer to inverse oflaplace 3/((s^2+9)(s^2+1)) is -1/8 sin(3t)+3/8 sin(t)