What is inverse oflaplace ((s-3))/((s^2+4s+13)) ?

The answer to inverse oflaplace ((s-3))/((s^2+4s+13)) is e^{-2t}cos(3t)-5/3 e^{-2t}sin(3t)

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inverse oflaplace ((s-3))/((s^2+4s+13))

inverse laplace

\frac{( s - 3 )}{( s ^{2} + 4 s + 13 )}

e^{- 2 t} cos (3 t) - \frac{5}{3} e^{- 2 t} sin (3 t)

Solution steps

L^{- 1} {\frac{( s - 3 )}{( s ^{2} + 4 s + 13 )}}

Expand

\frac{s - 3}{s ^{2} + 4 s + 13} : \frac{s + 2}{( s + 2 ) ^{2} + 9} - 5 \cdot \frac{1}{( s + 2 ) ^{2} + 9}

= L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 9} - 5 \cdot \frac{1}{( s + 2 ) ^{2} + 9}}

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} + 9}} - 5 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 9}}

L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 9}} : e^{- 2 t} cos (3 t)

L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 9}} : e^{- 2 t} \frac{1}{3} sin (3 t)

= e^{- 2 t} cos (3 t) - 5 e^{- 2 t} \frac{1}{3} sin (3 t)

Refine

e^{- 2 t} cos (3 t) - 5 e^{- 2 t} \frac{1}{3} sin (3 t) : e^{- 2 t} cos (3 t) - \frac{5}{3} e^{- 2 t} sin (3 t)

= e^{- 2 t} cos (3 t) - \frac{5}{3} e^{- 2 t} sin (3 t)

y^{^{'}} - 7 y = - x

y^{^{'}} = \frac{3100}{1 + 0.25 t}

\int (5 x + 4)^{5} d x

\frac{d}{d x} ((3 x^{4} - 1) (x - 1))

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = a, y^{^{'}} (0) = 5

What is inverse oflaplace ((s-3))/((s^2+4s+13)) ?
The answer to inverse oflaplace ((s-3))/((s^2+4s+13)) is e^{-2t}cos(3t)-5/3 e^{-2t}sin(3t)