What is inverse oflaplace (100)/(s^3+25s) ?

The answer to inverse oflaplace (100)/(s^3+25s) is 4\H(t)-4cos(5t)

Popular >

inverse oflaplace (100)/(s^3+25s)

inverse laplace

\frac{100}{s ^{3} + 25 s}

4 H (t) - 4 cos (5 t)

Solution steps

L^{- 1} {\frac{100}{s ^{3} + 25 s}}

Take the partial fraction of

\frac{100}{s ^{3} + 25 s} : \frac{4}{s} - \frac{4 s}{s ^{2} + 25}

= L^{- 1} {\frac{4}{s} - \frac{4 s}{s ^{2} + 25}}

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{4}{s}} - L^{- 1} {\frac{4 s}{s ^{2} + 25}}

L^{- 1} {\frac{4}{s}} : 4 H (t)

L^{- 1} {\frac{4 s}{s ^{2} + 25}} : 4 cos (5 t)

= 4 H (t) - 4 cos (5 t)

\frac{d}{d x} (x^{- \frac{6}{7}})

\frac{d}{d x} (\frac{6 x ^{2} + 2 x + 6}{x})

\frac{\partial}{\partial x} (7^{x^{2} + 1})

x x x

\int csc^{2} (4 x) d x

What is inverse oflaplace (100)/(s^3+25s) ?
The answer to inverse oflaplace (100)/(s^3+25s) is 4\H(t)-4cos(5t)