What is inverse oflaplace 8/(9s^2+9s+4) ?

The answer to inverse oflaplace 8/(9s^2+9s+4) is (16e^{-t/2}sin((sqrt(7)t)/6))/(3sqrt(7))

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inverse oflaplace 8/(9s^2+9s+4)

inverse laplace

\frac{8}{9 s ^{2} + 9 s + 4}

\frac{16 e ^{- \frac{t}{2}} sin ( \frac{7 t}{6} )}{3 7}

Solution steps

L^{- 1} {\frac{8}{9 s ^{2} + 9 s + 4}}

Expand

\frac{8}{9 s ^{2} + 9 s + 4} : \frac{8}{9} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}}

= L^{- 1} {\frac{8}{9} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{8}{9} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}}}

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{7}{36}}} : e^{- \frac{t}{2}} \frac{6}{7} sin (\frac{7 t}{6})

= \frac{8}{9} e^{- \frac{t}{2}} \frac{6}{7} sin (\frac{7 t}{6})

Refine

\frac{8}{9} e^{- \frac{t}{2}} \frac{6}{7} sin (\frac{7 t}{6}) : \frac{16 e ^{- \frac{t}{2}} sin ( \frac{7 t}{6} )}{3 7}

= \frac{16 e ^{- \frac{t}{2}} sin ( \frac{7 t}{6} )}{3 7}

\frac{x ^{2} + 32 x}{8}

\frac{8}{s - 1} - \frac{3}{s}

(2 x^{2}) + (x - 8)

x \to 5 lim (\frac{x + 5}{x - 5})

x \to 0 lim (\frac{x}{x ^{2} + 4 x})

What is inverse oflaplace 8/(9s^2+9s+4) ?
The answer to inverse oflaplace 8/(9s^2+9s+4) is (16e^{-t/2}sin((sqrt(7)t)/6))/(3sqrt(7))