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

The answer to inverse oflaplace (2s^2+1)/(s+3) is 2\delta'(t)-6\delta(t)+19e^{-3t}

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

inverse laplace

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

2 δ' (t) - 6 δ (t) + 19 e^{- 3 t}

Solution steps

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

Take the partial fraction of

\frac{2 s ^{2} + 1}{s + 3} : 2 s - 6 + \frac{19}{s + 3}

= L^{- 1} {2 s - 6 + \frac{19}{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)}

= 2 L^{- 1} {s} - L^{- 1} {6} + L^{- 1} {\frac{19}{s + 3}}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {6} : 6 δ (t)

L^{- 1} {\frac{19}{s + 3}} : 19 e^{- 3 t}

= 2 δ^{^{'}} (t) - 6 δ (t) + 19 e^{- 3 t}

\frac{x d y}{d x} = y + x e^{- \frac{y}{x}}

\frac{d}{d x} ((ln (x) + 3 x)^{2})

\frac{\partial}{\partial x} (ln (x + 9 y + 2 z))

\frac{d}{d x} (2 tan (x + \frac{π}{12}))

\frac{d}{d x} ((x + 1)^{\frac{2}{3}} (3 x^{2} - 2)^{3})

What is inverse oflaplace (2s^2+1)/(s+3) ?
The answer to inverse oflaplace (2s^2+1)/(s+3) is 2\delta'(t)-6\delta(t)+19e^{-3t}