What is inverse oflaplace s/((s^2+1)(s^2+4)) ?

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

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inverse oflaplace s/((s^2+1)(s^2+4))

inverse laplace

\frac{s}{( s ^{2} + 1 ) ( s ^{2} + 4 )}

\frac{1}{3} cos (t) - \frac{1}{3} cos (2 t)

Solution steps

L^{- 1} {\frac{s}{( s ^{2} + 1 ) ( s ^{2} + 4 )}}

Take the partial fraction of

\frac{s}{( s ^{2} + 1 ) ( s ^{2} + 4 )} : \frac{s}{3 ( s ^{2} + 1 )} - \frac{s}{3 ( s ^{2} + 4 )}

= L^{- 1} {\frac{s}{3 ( s ^{2} + 1 )} - \frac{s}{3 ( s ^{2} + 4 )}}

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}{3 ( s ^{2} + 1 )}} - L^{- 1} {\frac{s}{3 ( s ^{2} + 4 )}}

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

L^{- 1} {\frac{s}{3 ( s ^{2} + 4 )}} : \frac{1}{3} cos (2 t)

= \frac{1}{3} cos (t) - \frac{1}{3} cos (2 t)

x \to 1 lim (2 x^{3} - 5 x^{2} + x + 9)

\int \frac{3 x ^{2} + 2 x - 10}{x ^{3} + x ^{2} - 10 x + 8} d x

\int \frac{1}{x ln ( x )} d x

\int_{0}^{1} \frac{2 x + 3}{( x + 1 ) ^{2}} d x

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

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