What is inverse oflaplace (400)/(s(s^2+200)) ?

The answer to inverse oflaplace (400)/(s(s^2+200)) is 2\H(t)-2cos(10sqrt(2)t)

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inverse oflaplace (400)/(s(s^2+200))

inverse laplace

\frac{400}{s ( s ^{2} + 200 )}

2 H (t) - 2 cos (102 t)

Solution steps

L^{- 1} {\frac{400}{s ( s ^{2} + 200 )}}

Take the partial fraction of

\frac{400}{s ( s ^{2} + 200 )} : \frac{2}{s} - \frac{2 s}{s ^{2} + 200}

= L^{- 1} {\frac{2}{s} - \frac{2 s}{s ^{2} + 200}}

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{2}{s}} - L^{- 1} {\frac{2 s}{s ^{2} + 200}}

L^{- 1} {\frac{2}{s}} : 2 H (t)

L^{- 1} {\frac{2 s}{s ^{2} + 200}} : 2 cos (102 t)

= 2 H (t) - 2 cos (102 t)

x \to \infty lim (3^{- x})

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

\frac{\partial}{\partial x} (3 sin (4 x))

\frac{d}{d x} (- \frac{x ^{4}}{4})

\frac{d y}{y ^{3}} = x^{- 2} d x

What is inverse oflaplace (400)/(s(s^2+200)) ?
The answer to inverse oflaplace (400)/(s(s^2+200)) is 2\H(t)-2cos(10sqrt(2)t)