What is inverse oflaplace (3s)/(s^2-2s+2) ?

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

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inverse oflaplace (3s)/(s^2-2s+2)

inverse laplace

\frac{3 s}{s ^{2} - 2 s + 2}

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

Solution steps

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

Expand

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

= L^{- 1} {3 \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)}

= 3 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)

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

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

\int cos^{5} (9 - 11 x) d x

\frac{d}{d x} (\frac{3 x - 4}{2 x - 8})

\int 8 x - 5 d x

\frac{1}{3 ( 1 + x ^{2} )}, \frac{1}{6} x^{2}

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