What is inverse oflaplace (4x-1)/(2x+3) ?

The answer to inverse oflaplace (4x-1)/(2x+3) is 2\delta(t)-7/2 e^{-(3t)/2}

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inverse oflaplace (4x-1)/(2x+3)

inverse laplace

\frac{4 x - 1}{2 x + 3}

2 δ (t) - \frac{7}{2} e^{- \frac{3 t}{2}}

Solution steps

L^{- 1} {\frac{4 x - 1}{2 x + 3}}

Take the partial fraction of

\frac{4 x - 1}{2 x + 3} : 2 - \frac{7}{2 x + 3}

= L^{- 1} {2 - \frac{7}{2 x + 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} {2} - L^{- 1} {\frac{7}{2 x + 3}}

L^{- 1} {2} : 2 δ (t)

L^{- 1} {\frac{7}{2 x + 3}} : \frac{7}{2} e^{- \frac{3 t}{2}}

= 2 δ (t) - \frac{7}{2} e^{- \frac{3 t}{2}}

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What is inverse oflaplace (4x-1)/(2x+3) ?
The answer to inverse oflaplace (4x-1)/(2x+3) is 2\delta(t)-7/2 e^{-(3t)/2}