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

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

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

inverse laplace

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

e^{t} + e^{- 3 t}

Solution steps

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

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

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

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

f (x) = x^{3} + x^{2} - x - 3, at x = - 2

3 x^{2} - 4 x + 5

f (x) = 3 sin (x) - 33 cos (x)

6 + sec (x) sin (x)

2 x^{2} + 2 x

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