What is inverse oflaplace 1/(s(s^2+36)) ?

The answer to inverse oflaplace 1/(s(s^2+36)) is 1/36 \H(t)-1/36 cos(6t)

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inverse oflaplace 1/(s(s^2+36))

inverse laplace

\frac{1}{s ( s ^{2} + 36 )}

\frac{1}{36} H (t) - \frac{1}{36} cos (6 t)

Solution steps

L^{- 1} {\frac{1}{s ( s ^{2} + 36 )}}

Take the partial fraction of

\frac{1}{s ( s ^{2} + 36 )} : \frac{1}{36 s} - \frac{s}{36 ( s ^{2} + 36 )}

= L^{- 1} {\frac{1}{36 s} - \frac{s}{36 ( s ^{2} + 36 )}}

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{1}{36 s}} - L^{- 1} {\frac{s}{36 ( s ^{2} + 36 )}}

L^{- 1} {\frac{1}{36 s}} : \frac{1}{36} H (t)

L^{- 1} {\frac{s}{36 ( s ^{2} + 36 )}} : \frac{1}{36} cos (6 t)

= \frac{1}{36} H (t) - \frac{1}{36} cos (6 t)

(4 x + 4 y)^{3} = 64 x^{3} + 64 y^{3}, (- 1, 1)

\frac{d}{d x} (- \frac{49 x}{36 y})

7 x + 4, at x = 3

\frac{d}{d x} (- e^{5 x})

\frac{\partial}{\partial y} (x^{2} + y^{2} + x^{2} y + 4)

What is inverse oflaplace 1/(s(s^2+36)) ?
The answer to inverse oflaplace 1/(s(s^2+36)) is 1/36 \H(t)-1/36 cos(6t)