What is inverse oflaplace ((s-5))/(s^2+4s+20) ?

The answer to inverse oflaplace ((s-5))/(s^2+4s+20) is e^{-2t}cos(4t)-7/4 e^{-2t}sin(4t)

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inverse oflaplace ((s-5))/(s^2+4s+20)

inverse laplace

\frac{( s - 5 )}{s ^{2} + 4 s + 20}

e^{- 2 t} cos (4 t) - \frac{7}{4} e^{- 2 t} sin (4 t)

Solution steps

L^{- 1} {\frac{( s - 5 )}{s ^{2} + 4 s + 20}}

Expand

\frac{s - 5}{s ^{2} + 4 s + 20} : \frac{s + 2}{( s + 2 ) ^{2} + 16} - 7 \cdot \frac{1}{( s + 2 ) ^{2} + 16}

= L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 16} - 7 \cdot \frac{1}{( s + 2 ) ^{2} + 16}}

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{s + 2}{( s + 2 ) ^{2} + 16}} - 7 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 16}}

L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 16}} : e^{- 2 t} cos (4 t)

L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 16}} : e^{- 2 t} \frac{1}{4} sin (4 t)

= e^{- 2 t} cos (4 t) - 7 e^{- 2 t} \frac{1}{4} sin (4 t)

Refine

e^{- 2 t} cos (4 t) - 7 e^{- 2 t} \frac{1}{4} sin (4 t) : e^{- 2 t} cos (4 t) - \frac{7}{4} e^{- 2 t} sin (4 t)

= e^{- 2 t} cos (4 t) - \frac{7}{4} e^{- 2 t} sin (4 t)

24 y^{^{''}} + 13 y^{^{'}} - 7 y = 0

\int \frac{1}{( 1 + y ^{2} ) ^{\frac{3}{2}}} d y

f (x) = x ln (x + 1)

\frac{d}{d x} (sin (x) cos (x))

\frac{\partial}{\partial x} (\frac{7 x y}{x ^{2} + y ^{4}})

What is inverse oflaplace ((s-5))/(s^2+4s+20) ?
The answer to inverse oflaplace ((s-5))/(s^2+4s+20) is e^{-2t}cos(4t)-7/4 e^{-2t}sin(4t)