What is inverse oflaplace ((s+5))/((s^2+4s+13)) ?

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

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inverse oflaplace ((s+5))/((s^2+4s+13))

inverse laplace

\frac{( s + 5 )}{( s ^{2} + 4 s + 13 )}

e^{- 2 t} cos (3 t) + e^{- 2 t} sin (3 t)

Solution steps

L^{- 1} {\frac{( s + 5 )}{( s ^{2} + 4 s + 13 )}}

Expand

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

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

Refine

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

= e^{- 2 t} cos (3 t) + e^{- 2 t} sin (3 t)

\frac{d}{d x} (x^{2} (1 - x)^{3})

\int \frac{( tan ^{3} ( ln ( x ) ) )}{6 x} d x

9 y^{^{''}} + 10 y = 0

\frac{\partial}{\partial v} (u)

\int_{1}^{2} (x^{2} + \frac{1}{4 x ^{2}}) d x

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