What is inverse oflaplace 1/(s+3)-1/(s+4) ?

The answer to inverse oflaplace 1/(s+3)-1/(s+4) is e^{-3t}-e^{-4t}

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inverse oflaplace 1/(s+3)-1/(s+4)

inverse laplace

\frac{1}{s + 3} - \frac{1}{s + 4}

e^{- 3 t} - e^{- 4 t}

Solution steps

L^{- 1} {\frac{1}{s + 3} - \frac{1}{s + 4}}

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 + 3}} - L^{- 1} {\frac{1}{s + 4}}

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

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

= e^{- 3 t} - e^{- 4 t}

\frac{\partial}{\partial s} (\frac{( 2 r + s )}{t})

\frac{d}{d y} (x sin (x y))

x \to 1 lim (\frac{1}{x - 2})

\frac{\partial}{\partial x} (3 y sin (7 y^{2} x))

\frac{2 x + 8 x ^{2}}{( 1 + 8 x ) ^{2}}

What is inverse oflaplace 1/(s+3)-1/(s+4) ?
The answer to inverse oflaplace 1/(s+3)-1/(s+4) is e^{-3t}-e^{-4t}