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} (\frac{( x + 3 ) ( x + 1 )}{x ^{2} - 49})

\int_{0}^{1} x cos (π x) d x

\int_{1}^{\infty} x^{- \frac{3}{2}} d x

y^{^{''}} - \frac{2}{x} y^{^{'}} + \frac{2}{x ^{2}} y = 0

\frac{\partial}{\partial x} (sec^{2} (x) - 1)

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)