What is inverse oflaplace (s+1)/((s^2-2s+5)) ?

The answer to inverse oflaplace (s+1)/((s^2-2s+5)) is e^tcos(2t)+e^tsin(2t)

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inverse oflaplace (s+1)/((s^2-2s+5))

inverse laplace

\frac{s + 1}{( s ^{2} - 2 s + 5 )}

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

Solution steps

L^{- 1} {\frac{s + 1}{( s ^{2} - 2 s + 5 )}}

Expand

\frac{s + 1}{s ^{2} - 2 s + 5} : \frac{s - 1}{( s - 1 ) ^{2} + 4} + 2 \cdot \frac{1}{( s - 1 ) ^{2} + 4}

= L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 4} + 2 \cdot \frac{1}{( s - 1 ) ^{2} + 4}}

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} + 4}} + 2 L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 4}}

L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 4}} : e^{t} cos (2 t)

L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 4}} : e^{t} \frac{1}{2} sin (2 t)

= e^{t} cos (2 t) + 2 e^{t} \frac{1}{2} sin (2 t)

Refine

e^{t} cos (2 t) + 2 e^{t} \frac{1}{2} sin (2 t) : e^{t} cos (2 t) + e^{t} sin (2 t)

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

\frac{\partial}{\partial x} (x yz + x y^{5} + y z^{5} + z x^{5})

y = x^{2} - 3, y = 1

\frac{d}{d x} (x^{3} - 12 x y + 8 y^{3})

\int (x - 1) (x + 2) d x

What is inverse oflaplace (s+1)/((s^2-2s+5)) ?
The answer to inverse oflaplace (s+1)/((s^2-2s+5)) is e^tcos(2t)+e^tsin(2t)