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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

= L^{- 1} {\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)}

= 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)

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

Refine

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

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

\int_{1}^{5} 2 e^{2 x} d x

x \to - 0.001 lim (4 x^{2} + 7)

\frac{\partial}{\partial x} (4 x^{3} y^{3} - 18 y - x)

\frac{\partial}{\partial y} (9 x sin (y - z))

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