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

The answer to inverse oflaplace ((3s+1))/(s+2) is 3\delta(t)-5e^{-2t}

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

inverse laplace

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

3 δ (t) - 5 e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{3 s + 1}{s + 2} : 3 - \frac{5}{s + 2}

= L^{- 1} {3 - \frac{5}{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} {3} - L^{- 1} {\frac{5}{s + 2}}

L^{- 1} {3} : 3 δ (t)

L^{- 1} {\frac{5}{s + 2}} : 5 e^{- 2 t}

= 3 δ (t) - 5 e^{- 2 t}

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

\int_{1}^{3} (x^{3} + \frac{1}{4 x ^{3}}) d x

y = 2 x - 1 + 4 x^{2}, (1, 5)

\int k e^{- x} d x

\frac{\partial}{\partial x} (y (e^{x} + x e^{x}))

What is inverse oflaplace ((3s+1))/(s+2) ?
The answer to inverse oflaplace ((3s+1))/(s+2) is 3\delta(t)-5e^{-2t}