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

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

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

inverse laplace

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

δ (t) - e^{- 3 t}

Solution steps

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

Take the partial fraction of

\frac{s + 2}{s + 3} : 1 - \frac{1}{s + 3}

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

L^{- 1} {1} : δ (t)

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

= δ (t) - e^{- 3 t}

\frac{d}{d x} (x^{5} + 5 x^{4} - 10 x^{2} + 6)

\frac{d y}{d x}, 2 x^{4} + 9 y = 2 x^{2} y + 36

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

\frac{( s )}{s ^{2} + 5 s + 6}

\frac{d}{d x} (\frac{( - 2 x - 3 ) ^{7}}{( 4 x + 2 ) ^{4}})

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