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

The answer to inverse oflaplace ((-2))/((6s+s^2)) is -1/3 \H(t)+1/3 e^{-6t}

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inverse oflaplace ((-2))/((6s+s^2))

inverse laplace

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

- \frac{1}{3} H (t) + \frac{1}{3} e^{- 6 t}

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {- \frac{1}{3 s} + \frac{1}{3 ( s + 6 )}}

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}{3 s}} + L^{- 1} {\frac{1}{3 ( s + 6 )}}

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

L^{- 1} {\frac{1}{3 ( s + 6 )}} : \frac{1}{3} e^{- 6 t}

= - \frac{1}{3} H (t) + \frac{1}{3} e^{- 6 t}

y d x = 2 (x + y) d y

(x + y) d x + (x + 2 y) d y = 0, y (2) = 3

x \to 0 - lim (\frac{1}{1 + e ^{\frac{1}{x}}})

\frac{d}{d x} (2 arctan (4 x))

x \to 0 lim (x^{2} ln (x^{2}))

What is inverse oflaplace ((-2))/((6s+s^2)) ?
The answer to inverse oflaplace ((-2))/((6s+s^2)) is -1/3 \H(t)+1/3 e^{-6t}