What is inverse oflaplace 1/(s^2-2s+4) ?

The answer to inverse oflaplace 1/(s^2-2s+4) is e^t 1/(sqrt(3))sin(sqrt(3)t)

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inverse oflaplace 1/(s^2-2s+4)

inverse laplace

\frac{1}{s ^{2} - 2 s + 4}

e^{t} \frac{1}{3} sin (3 t)

Solution steps

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

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

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s - 1 ) ^{2} + 3} : a = 1, F (s) = \frac{1}{s ^{2} + 3}

= e^{t} L^{- 1} {\frac{1}{s ^{2} + 3}}

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

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

x \frac{d y}{d x} - 4 y = 0

e^{4 x}

\int \frac{1}{x ( 1 + 4 ln ^{2} ( x ) )} d x

\frac{d y}{d x} = 2 x + 2 x y

x \to \infty lim (3 x^{5})

What is inverse oflaplace 1/(s^2-2s+4) ?
The answer to inverse oflaplace 1/(s^2-2s+4) is e^t 1/(sqrt(3))sin(sqrt(3)t)