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

The answer to inverse oflaplace 9/((s^2+4s+9)) is 9/(sqrt(5))e^{-2t}sin(sqrt(5)t)

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

inverse laplace

\frac{9}{( s ^{2} + 4 s + 9 )}

\frac{9}{5} e^{- 2 t} sin (5 t)

Solution steps

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

Expand

\frac{9}{s ^{2} + 4 s + 9} : 9 \cdot \frac{1}{( s + 2 ) ^{2} + 5}

= L^{- 1} {9 \cdot \frac{1}{( s + 2 ) ^{2} + 5}}

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 + 2 ) ^{2} + 5}}

L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 5}} : e^{- 2 t} \frac{1}{5} sin (5 t)

= 9 e^{- 2 t} \frac{1}{5} sin (5 t)

Refine

9 e^{- 2 t} \frac{1}{5} sin (5 t) : \frac{9}{5} e^{- 2 t} sin (5 t)

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

y = 8 x^{4}, at x = 2

f (x) = (x^{3} - 4 x)^{8}, at x = 2

f (x) = 3^{2 x}

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

\int 2 x^{2} 1 - x^{3} d x

What is inverse oflaplace 9/((s^2+4s+9)) ?
The answer to inverse oflaplace 9/((s^2+4s+9)) is 9/(sqrt(5))e^{-2t}sin(sqrt(5)t)