What is inverse oflaplace (s+1)/(s^2+3s) ?

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

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inverse oflaplace (s+1)/(s^2+3s)

inverse laplace

\frac{s + 1}{s ^{2} + 3 s}

\frac{1}{3} H (t) + \frac{2}{3} e^{- 3 t}

Solution steps

L^{- 1} {\frac{s + 1}{s ^{2} + 3 s}}

Take the partial fraction of

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

= L^{- 1} {\frac{1}{3 s} + \frac{2}{3 ( s + 3 )}}

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{2}{3 ( s + 3 )}}

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

L^{- 1} {\frac{2}{3 ( s + 3 )}} : \frac{2}{3} e^{- 3 t}

= \frac{1}{3} H (t) + \frac{2}{3} e^{- 3 t}

x \cdot \frac{d y}{d x} - y = 3 x^{6} y, y (1) = 2

\int_{2}^{7} 2 π y (7 - y) d y

\int \frac{1}{9 - 7 x} d x

\frac{\partial}{\partial x} (\frac{1}{x ^{3} + 4 x ^{2}})

\frac{d}{d x} (cot (8 x))

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