What is inverse oflaplace ((2s+2))/((s+10)) ?

The answer to inverse oflaplace ((2s+2))/((s+10)) is 2\delta(t)-18e^{-10t}

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inverse oflaplace ((2s+2))/((s+10))

inverse laplace

\frac{( 2 s + 2 )}{( s + 10 )}

2 δ (t) - 18 e^{- 10 t}

Solution steps

L^{- 1} {\frac{( 2 s + 2 )}{( s + 10 )}}

Take the partial fraction of

\frac{2 s + 2}{s + 10} : 2 - \frac{18}{s + 10}

= L^{- 1} {2 - \frac{18}{s + 10}}

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} {2} - L^{- 1} {\frac{18}{s + 10}}

L^{- 1} {2} : 2 δ (t)

L^{- 1} {\frac{18}{s + 10}} : 18 e^{- 10 t}

= 2 δ (t) - 18 e^{- 10 t}

\int \frac{1}{P ( a - b P )} d P

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

\frac{\partial}{\partial x} (- sin (2 x))

y = \frac{( x ^{2} + 3 )}{2 x}

\frac{d}{d x} (ln (12 cos^{2} (x)))

What is inverse oflaplace ((2s+2))/((s+10)) ?
The answer to inverse oflaplace ((2s+2))/((s+10)) is 2\delta(t)-18e^{-10t}