What is inverse oflaplace 1000*1/(s^2+1000s) ?

The answer to inverse oflaplace 1000*1/(s^2+1000s) is 1000(1/1000 \H(t)-1/1000 e^{-1000t})

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inverse oflaplace 1000*1/(s^2+1000s)

inverse laplace

1000 \cdot \frac{1}{s ^{2} + 1000 s}

1000 (\frac{1}{1000} H (t) - \frac{1}{1000} e^{- 1000 t})

Solution steps

L^{- 1} {1000 \cdot \frac{1}{s ^{2} + 1000 s}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 1000 L^{- 1} {\frac{1}{s ^{2} + 1000 s}}

L^{- 1} {\frac{1}{s ^{2} + 1000 s}} : \frac{1}{1000} H (t) - \frac{1}{1000} e^{- 1000 t}

= 1000 (\frac{1}{1000} H (t) - \frac{1}{1000} e^{- 1000 t})

\int \frac{x ^{4}}{4} - \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x + C d x

\int 2 x^{3} d x

\frac{3}{ln ( x )}

x^{2} y^{^{''}} - x y^{^{'}} - 3 y = 0

\frac{x d y}{d x} = \frac{1}{y ^{3}}

What is inverse oflaplace 1000*1/(s^2+1000s) ?
The answer to inverse oflaplace 1000*1/(s^2+1000s) is 1000(1/1000 \H(t)-1/1000 e^{-1000t})