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

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

Popular >

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

inverse laplace

\frac{2}{s ^{2} + 2 s}

H (t) - e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{2}{s ^{2} + 2 s} : \frac{1}{s} - \frac{1}{s + 2}

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

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

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

L^{- 1} {\frac{1}{s + 2}} : e^{- 2 t}

= H (t) - e^{- 2 t}

\frac{d y}{d t} = \frac{9}{40} - \frac{y}{200}, y (0) = 90

\frac{d}{d t} (7 cos (t))

y = ln (1 + sin (x))

\int \frac{sin ( x )}{7 + cos ^{2} ( x )} d x

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