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

The answer to inverse oflaplace 1/(s^2-3s+5) is e^{(3t)/2} 2/(sqrt(11))sin((sqrt(11)t)/2)

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

inverse laplace

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

e^{\frac{3 t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

Solution steps

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

\frac{1}{s ^{2} - 3 s + 5} = \frac{1}{( s - \frac{3}{2} ) ^{2} + \frac{11}{4}}

= L^{- 1} {\frac{1}{( s - \frac{3}{2} ) ^{2} + \frac{11}{4}}}

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 - \frac{3}{2} ) ^{2} + \frac{11}{4}} : a = \frac{3}{2}, F (s) = \frac{1}{s ^{2} + \frac{11}{4}}

= e^{\frac{3}{2} t} L^{- 1} {\frac{1}{s ^{2} + \frac{11}{4}}}

L^{- 1} {\frac{1}{s ^{2} + \frac{11}{4}}} : \frac{2}{11} sin (\frac{11 t}{2})

= e^{\frac{3 t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

y^{^{'}} = \frac{( 4 t ^{3} + 1 )}{( 2 y - 6 )}

x \to - 1 lim (\frac{2 x}{∣ x ∣})

\int \frac{e ^{x^{\frac{1}{2}}}}{x ^{3}} d x

\int (- 2 T + x) \cdot e^{- i \cdot z \cdot x} d x

(t^{2}) \cdot (sin (2 \cdot t))

What is inverse oflaplace 1/(s^2-3s+5) ?
The answer to inverse oflaplace 1/(s^2-3s+5) is e^{(3t)/2} 2/(sqrt(11))sin((sqrt(11)t)/2)