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)

\int_{2}^{3} 3 x^{2} d x

2 x + 3

\frac{d}{d x} ((1 + 1 0^{l n (x)})^{6})

\int \frac{1}{3 - 2 x - x ^{2}} d x

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

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)