What is inverse oflaplace-4e^{(-3s)}*1/(s-3) ?

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

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inverse oflaplace-4e^{(-3s)}*1/(s-3)

inverse laplace

- 4 e^{(- 3 s)} \cdot \frac{1}{s - 3}

- 4 H (t - 3) e^{3 (t - 3)}

Solution steps

L^{- 1} {- 4 e^{(- 3 s)} \frac{1}{s - 3}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= - 4 L^{- 1} {e^{- 3 s} \frac{1}{s - 3}}

L^{- 1} {e^{- 3 s} \frac{1}{s - 3}} : H (t - 3) e^{3 (t - 3)}

= - 4 H (t - 3) e^{3 (t - 3)}

\frac{d}{d x} (2 + cos (e^{x} - 2) - e^{x})

\frac{\partial}{\partial x} (2^{90})

\int_{1}^{5} x^{4} ln (5 x) d x

y = x^{3 - 3 x + 1}

\int 4 x - y^{2} d y

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