What is inverse oflaplace (s+12)/((s^2+4s)) ?

The answer to inverse oflaplace (s+12)/((s^2+4s)) is 3\H(t)-2e^{-4t}

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inverse oflaplace (s+12)/((s^2+4s))

inverse laplace

\frac{s + 12}{( s ^{2} + 4 s )}

3 H (t) - 2 e^{- 4 t}

Solution steps

L^{- 1} {\frac{s + 12}{( s ^{2} + 4 s )}}

Take the partial fraction of

\frac{s + 12}{s ^{2} + 4 s} : \frac{3}{s} - \frac{2}{s + 4}

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

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

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

L^{- 1} {\frac{2}{s + 4}} : 2 e^{- 4 t}

= 3 H (t) - 2 e^{- 4 t}

\frac{d}{d t} (\frac{5}{( 5 - t ) ^{2}})

\int \frac{x - 5}{x ^{3} + 2 x ^{2} + x} d x

\frac{\partial}{\partial x} (x ln (z^{5} + x^{4}))

\frac{\partial}{\partial x} (7 x y^{2} + 1)

\frac{d}{d x} (4 ln (2 x))

What is inverse oflaplace (s+12)/((s^2+4s)) ?
The answer to inverse oflaplace (s+12)/((s^2+4s)) is 3\H(t)-2e^{-4t}