What is inverse oflaplace ((2s+1))/((s^2-2s+2)) ?

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

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inverse oflaplace ((2s+1))/((s^2-2s+2))

inverse laplace

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

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

Solution steps

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

Expand

\frac{2 s + 1}{s ^{2} - 2 s + 2} : 2 \cdot \frac{s - 1}{( s - 1 ) ^{2} + 1} + 3 \cdot \frac{1}{( s - 1 ) ^{2} + 1}

= L^{- 1} {2 \cdot \frac{s - 1}{( s - 1 ) ^{2} + 1} + 3 \cdot \frac{1}{( s - 1 ) ^{2} + 1}}

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)}

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

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

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

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

\frac{\partial}{\partial x} ((y^{2}) d x)

(2 x + 3)^{2}

4 x^{2} (\frac{d y}{d x}) = (\frac{d y}{d x}) + 3 x e^{- y}

\frac{d}{d x} (- x^{2} y^{2})

x \to 0 lim (x cos (\frac{14}{x}))

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