What is inverse oflaplace (3s-1)/((s^2+2s+5)) ?

The answer to inverse oflaplace (3s-1)/((s^2+2s+5)) is 3e^{-t}cos(2t)-2e^{-t}sin(2t)

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inverse oflaplace (3s-1)/((s^2+2s+5))

inverse laplace

\frac{3 s - 1}{( s ^{2} + 2 s + 5 )}

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

Solution steps

L^{- 1} {\frac{3 s - 1}{( s ^{2} + 2 s + 5 )}}

Expand

\frac{3 s - 1}{s ^{2} + 2 s + 5} : 3 \cdot \frac{s + 1}{( s + 1 ) ^{2} + 4} - 4 \cdot \frac{1}{( s + 1 ) ^{2} + 4}

= L^{- 1} {3 \cdot \frac{s + 1}{( s + 1 ) ^{2} + 4} - 4 \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)}

= 3 L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 4}} - 4 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)

= 3 e^{- t} cos (2 t) - 4 e^{- t} \frac{1}{2} sin (2 t)

Refine

3 e^{- t} cos (2 t) - 4 e^{- t} \frac{1}{2} sin (2 t) : 3 e^{- t} cos (2 t) - 2 e^{- t} sin (2 t)

= 3 e^{- t} cos (2 t) - 2 e^{- t} sin (2 t)

\frac{d y}{d x} = 3 (y^{2} + 1), y (\frac{π}{4}) = 1

y^{^{'}} - y = 2 t e^{2 t}

\frac{\partial}{\partial y} (2 x e^{x^{2} - y^{2}})

\int y e^{- y} d y

\frac{d}{d x} (cos^{2} (sin (4 x)))

What is inverse oflaplace (3s-1)/((s^2+2s+5)) ?
The answer to inverse oflaplace (3s-1)/((s^2+2s+5)) is 3e^{-t}cos(2t)-2e^{-t}sin(2t)