What is inverse oflaplace s/(s^2-2s+10) ?

The answer to inverse oflaplace s/(s^2-2s+10) is e^tcos(3t)+1/3 e^tsin(3t)

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inverse oflaplace s/(s^2-2s+10)

inverse laplace

\frac{s}{s ^{2} - 2 s + 10}

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

Solution steps

L^{- 1} {\frac{s}{s ^{2} - 2 s + 10}}

Expand

\frac{s}{s ^{2} - 2 s + 10} : \frac{s - 1}{( s - 1 ) ^{2} + 9} + 1 \cdot \frac{1}{( s - 1 ) ^{2} + 9}

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

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

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

= e^{t} cos (3 t) + 1 \cdot e^{t} \frac{1}{3} sin (3 t)

Refine

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

= e^{t} cos (3 t) + \frac{1}{3} e^{t} sin (3 t)

\frac{d}{d x} (\frac{6 x ^{2} + 4 x + 2}{x})

\frac{d}{d y} (\frac{y ^{4} + 48}{24 y})

\int \frac{x ^{3} - 1}{x ^{4} - 4 x} d x

\frac{\partial}{\partial x} (\frac{1}{\frac{1}{x} + \frac{1}{y}})

\frac{\partial}{\partial x} (4 arctan (x y))

What is inverse oflaplace s/(s^2-2s+10) ?
The answer to inverse oflaplace s/(s^2-2s+10) is e^tcos(3t)+1/3 e^tsin(3t)