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

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

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

inverse laplace

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

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

Solution steps

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

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

L^{- 1} {\frac{2}{s}} : 2 H (t)

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

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

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

\frac{( e ^{t} )}{1 + t}

x^{2} + x, at x = 1, x = 4

\frac{d}{d x} (\frac{24}{( x + 1 ) ^{5}})

\frac{1}{x}

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