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

The answer to inverse oflaplace (2s+1)/(s^2+2s+2) is 2e^{-t}cos(t)-e^{-t}sin(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) - 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} - 1 \cdot \frac{1}{( s + 1 ) ^{2} + 1}

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

Refine

2 e^{- t} cos (t) - 1 e^{- t} sin (t) : 2 e^{- t} cos (t) - e^{- t} sin (t)

= 2 e^{- t} cos (t) - e^{- t} sin (t)

\frac{\partial}{\partial z} (cos (y) sin (x))

ln (x y)

y = x^{3} - 13 x, (1, - 12)

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 11

x^{6} + y^{5} x^{7} + 5 y^{^{'}} = 0

What is inverse oflaplace (2s+1)/(s^2+2s+2) ?
The answer to inverse oflaplace (2s+1)/(s^2+2s+2) is 2e^{-t}cos(t)-e^{-t}sin(t)