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

The answer to inverse oflaplace 5/((s^2+2s+2)) is 5e^{-t}sin(t)

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

inverse laplace

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

5 e^{- t} sin (t)

Solution steps

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

Expand

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

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

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

= 5 e^{- t} sin (t)

\frac{\partial}{\partial x} (\frac{( 4 x - y )}{4 x + y})

\frac{d}{d x} (100 x e^{- 0.5 x})

\frac{\partial}{\partial y} (6 x)

f (x) = x - e^{- x}

\frac{\partial}{\partial x} (x^{2} - 3 x y + 6 y^{2} + x - 2 y)

What is inverse oflaplace 5/((s^2+2s+2)) ?
The answer to inverse oflaplace 5/((s^2+2s+2)) is 5e^{-t}sin(t)