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

The answer to inverse oflaplace 9/(s^2+9s+9) is 6/(sqrt(5))e^{-(9t)/2}sinh((3sqrt(5)t)/2)

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

inverse laplace

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

\frac{6}{5} e^{- \frac{9 t}{2}} sinh (\frac{3 5 t}{2})

Solution steps

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

Expand

\frac{9}{s ^{2} + 9 s + 9} : 9 \cdot \frac{1}{( s + \frac{9}{2} ) ^{2} - \frac{45}{4}}

= L^{- 1} {9 \cdot \frac{1}{( s + \frac{9}{2} ) ^{2} - \frac{45}{4}}}

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)}

= 9 L^{- 1} {\frac{1}{( s + \frac{9}{2} ) ^{2} - \frac{45}{4}}}

L^{- 1} {\frac{1}{( s + \frac{9}{2} ) ^{2} - \frac{45}{4}}} : e^{- \frac{9 t}{2}} \frac{2}{3 5} sinh (\frac{3 5 t}{2})

= 9 e^{- \frac{9 t}{2}} \frac{2}{3 5} sinh (\frac{3 5 t}{2})

Refine

9 e^{- \frac{9 t}{2}} \frac{2}{3 5} sinh (\frac{3 5 t}{2}) : \frac{6}{5} e^{- \frac{9 t}{2}} sinh (\frac{3 5 t}{2})

= \frac{6}{5} e^{- \frac{9 t}{2}} sinh (\frac{3 5 t}{2})

(e^{- ∣ x ∣})^{^{'}}

\int sin (x^{- 1}) d x

\int \frac{e ^{x}}{9 + e ^{x}} d x

\int \frac{5 x + 5}{x ^{2} + 1} d x

e^{3 x^{4}}

What is inverse oflaplace 9/(s^2+9s+9) ?
The answer to inverse oflaplace 9/(s^2+9s+9) is 6/(sqrt(5))e^{-(9t)/2}sinh((3sqrt(5)t)/2)