What is inverse oflaplace ((s+4))/((s^2-2s+4)) ?

The answer to inverse oflaplace ((s+4))/((s^2-2s+4)) is e^tcos(sqrt(3)t)+5/(sqrt(3))e^tsin(sqrt(3)t)

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inverse oflaplace ((s+4))/((s^2-2s+4))

inverse laplace

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

e^{t} cos (3 t) + \frac{5}{3} e^{t} sin (3 t)

Solution steps

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

Expand

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

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

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

L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 3}} : e^{t} cos (3 t)

L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 3}} : e^{t} \frac{1}{3} sin (3 t)

= e^{t} cos (3 t) + 5 e^{t} \frac{1}{3} sin (3 t)

Refine

e^{t} cos (3 t) + 5 e^{t} \frac{1}{3} sin (3 t) : e^{t} cos (3 t) + \frac{5}{3} e^{t} sin (3 t)

= e^{t} cos (3 t) + \frac{5}{3} e^{t} sin (3 t)

(- 16 x y^{2} + 3 y) + (- 16 x^{2} y + 3 x) y^{^{'}} = 0

x^{2} + x y + 2 y^{2} = 18, (3, - 3)

\int \frac{x}{x ^{2} + 6 x + 13} d x

\int x e^{- x} sin (x) d x

x \to - \infty lim (x^{4} e^{x})

What is inverse oflaplace ((s+4))/((s^2-2s+4)) ?
The answer to inverse oflaplace ((s+4))/((s^2-2s+4)) is e^tcos(sqrt(3)t)+5/(sqrt(3))e^tsin(sqrt(3)t)