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)

\frac{[ y - y ^{2} ]}{t} = y^{^{'}}, y (1) = 2

x \to 0 lim (x (11 + \frac{4}{x}))

\int cos (\frac{nπ}{2} x) d x

\frac{\partial}{\partial x} (\frac{2 x}{x ^{2} - 2 x})

x \to 1 lim (sin (\frac{1}{x}))

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)