What is inverse oflaplace ((12+4x))/(x-1) ?

The answer to inverse oflaplace ((12+4x))/(x-1) is 4\delta(t)+16e^t

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inverse oflaplace ((12+4x))/(x-1)

inverse laplace

\frac{( 12 + 4 x )}{x - 1}

4 δ (t) + 16 e^{t}

Solution steps

L^{- 1} {\frac{( 12 + 4 x )}{x - 1}}

Take the partial fraction of

\frac{12 + 4 x}{x - 1} : 4 + \frac{16}{x - 1}

= L^{- 1} {4 + \frac{16}{x - 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} {4} + L^{- 1} {\frac{16}{x - 1}}

L^{- 1} {4} : 4 δ (t)

L^{- 1} {\frac{16}{x - 1}} : 16 e^{t}

= 4 δ (t) + 16 e^{t}

n = 0 \sum \infty n (\frac{2}{7})^{n}

\frac{\partial}{\partial x} ((x^{3} + y^{2})^{5})

\int_{0}^{\frac{π}{2}} 20 cos^{5} (x) d x

\int_{1}^{2} (\frac{6}{x ^{3}} + 2 x) d x

x \to 2 lim (x^{3} - x^{2} + 2 x)

What is inverse oflaplace ((12+4x))/(x-1) ?
The answer to inverse oflaplace ((12+4x))/(x-1) is 4\delta(t)+16e^t