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

The answer to inverse oflaplace (s+1)/(s^2(s+2)) is 1/4 \H(t)+t/2-1/4 e^{-2t}

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

inverse laplace

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

\frac{1}{4} H (t) + \frac{t}{2} - \frac{1}{4} e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{s + 1}{s ^{2} ( s + 2 )} : \frac{1}{4 s} + \frac{1}{2 s ^{2}} - \frac{1}{4 ( s + 2 )}

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

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

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

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

= \frac{1}{4} H (t) + \frac{t}{2} - \frac{1}{4} e^{- 2 t}

\frac{d ^{2} y}{d x ^{2}} = y

\int - cos (x) ln (x) d x

\int \frac{x}{( 36 - x ^{2} ) ^{3}} d x

x \to \infty lim (1 + x^{2})

(e^{s i n (x)})^{^{'}}

What is inverse oflaplace (s+1)/(s^2(s+2)) ?
The answer to inverse oflaplace (s+1)/(s^2(s+2)) is 1/4 \H(t)+t/2-1/4 e^{-2t}