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

The answer to inverse oflaplace ((2s+3))/(s^2-2s+5) is 2e^tcos(2t)+5/2 e^tsin(2t)

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

inverse laplace

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

2 e^{t} cos (2 t) + \frac{5}{2} e^{t} sin (2 t)

Solution steps

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

Expand

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

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

= 2 L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 4}} + 5 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)

= 2 e^{t} cos (2 t) + 5 e^{t} \frac{1}{2} sin (2 t)

Refine

2 e^{t} cos (2 t) + 5 e^{t} \frac{1}{2} sin (2 t) : 2 e^{t} cos (2 t) + \frac{5}{2} e^{t} sin (2 t)

= 2 e^{t} cos (2 t) + \frac{5}{2} e^{t} sin (2 t)

y^{^{''}} - 6 y^{^{'}} - 27 y = - 3 e^{- 3 t}

\int (\frac{1}{x} - \frac{1}{e ^{2 x}}) d x

\int y^{2} e^{y^{2}} d y

n = 1 \sum \infty (\frac{1}{10})^{n}

\frac{d}{d x} ((x + 1)^{e^{x}})

What is inverse oflaplace ((2s+3))/(s^2-2s+5) ?
The answer to inverse oflaplace ((2s+3))/(s^2-2s+5) is 2e^tcos(2t)+5/2 e^tsin(2t)