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)

sin^{2} (x) + cos^{2} (x)

x \to 0.5 lim (\frac{sin ( x )}{x})

(- 5, - 1), (5, 2)

\int cos^{2} (\frac{3 π x}{L}) d x

n = 0 \sum \infty (- \frac{3}{10})^{n}

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)