What is inverse oflaplace ((10/6))/(s(s^2+1/2)) ?

The answer to inverse oflaplace ((10/6))/(s(s^2+1/2)) is 10/3 \H(t)-10/3 cos(sqrt(1/2)t)

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inverse oflaplace ((10/6))/(s(s^2+1/2))

inverse laplace

\frac{( \frac{10}{6} )}{s ( s ^{2} + \frac{1}{2} )}

\frac{10}{3} H (t) - \frac{10}{3} cos (\frac{1}{2} t)

Solution steps

L^{- 1} {\frac{( \frac{10}{6} )}{s ( s ^{2} + \frac{1}{2} )}}

Take the partial fraction of

\frac{\frac{10}{6}}{s ( s ^{2} + \frac{1}{2} )} : \frac{10}{3 s} - \frac{20 s}{3 ( 2 s ^{2} + 1 )}

= L^{- 1} {\frac{10}{3 s} - \frac{20 s}{3 ( 2 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{10}{3 s}} - L^{- 1} {\frac{20 s}{3 ( 2 s ^{2} + 1 )}}

L^{- 1} {\frac{10}{3 s}} : \frac{10}{3} H (t)

L^{- 1} {\frac{20 s}{3 ( 2 s ^{2} + 1 )}} : \frac{10}{3} cos (\frac{1}{2} t)

= \frac{10}{3} H (t) - \frac{10}{3} cos (\frac{1}{2} t)

\frac{5}{6 x ^{5}}

\int (- 3 x^{4} - 6 x^{2} + 8) d x

2 x \frac{d y}{d x} + 2 y = x^{3} y^{4}

\int 18 y^{2} d y

\int - \frac{4}{x ^{3}} d x

What is inverse oflaplace ((10/6))/(s(s^2+1/2)) ?
The answer to inverse oflaplace ((10/6))/(s(s^2+1/2)) is 10/3 \H(t)-10/3 cos(sqrt(1/2)t)