What is inverse oflaplace ((s+3))/((s+7)) ?

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

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inverse oflaplace ((s+3))/((s+7))

inverse laplace

\frac{( s + 3 )}{( s + 7 )}

δ (t) - 4 e^{- 7 t}

Solution steps

L^{- 1} {\frac{( s + 3 )}{( s + 7 )}}

Take the partial fraction of

\frac{s + 3}{s + 7} : 1 - \frac{4}{s + 7}

= L^{- 1} {1 - \frac{4}{s + 7}}

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} {1} - L^{- 1} {\frac{4}{s + 7}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{4}{s + 7}} : 4 e^{- 7 t}

= δ (t) - 4 e^{- 7 t}

y^{^{'}} + \frac{2 y}{t} = \frac{2 cos ( t ^{2} )}{t}

\int x \cdot ln (x + 1) d x

3 sin (3 x)

\int \frac{x ^{4} - 6 x ^{3} + 12 x ^{2} + 6}{x ^{3} - 6 x ^{2} + 12 x - 8} d x

\frac{d}{d x} (\frac{( x + 4 ) ( x ^{2} - 4 x + 16 )}{x ^{3}})

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