What is inverse oflaplace (0.9)/(0.1s^2+s+0.9) ?

The answer to inverse oflaplace (0.9)/(0.1s^2+s+0.9) is 9/4 e^{-5t}sinh(4t)

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inverse oflaplace (0.9)/(0.1s^2+s+0.9)

inverse laplace

\frac{0.9}{0.1 s ^{2} + s + 0.9}

\frac{9}{4} e^{- 5 t} sinh (4 t)

Solution steps

L^{- 1} {\frac{0.9}{0.1 s ^{2} + s + 0.9}}

Expand

\frac{0.9}{0.1 s ^{2} + s + 0.9} : 9 \cdot \frac{1}{( s + 5 ) ^{2} - 16}

= L^{- 1} {9 \cdot \frac{1}{( s + 5 ) ^{2} - 16}}

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 + 5 ) ^{2} - 16}}

L^{- 1} {\frac{1}{( s + 5 ) ^{2} - 16}} : e^{- 5 t} \frac{1}{4} sinh (4 t)

= 9 e^{- 5 t} \frac{1}{4} sinh (4 t)

Refine

9 e^{- 5 t} \frac{1}{4} sinh (4 t) : \frac{9}{4} e^{- 5 t} sinh (4 t)

= \frac{9}{4} e^{- 5 t} sinh (4 t)

\int \frac{5 + x}{x} d x

\int csc^{2} (x) - 1 d x

\int e^{(1 + n) x} d x

\int \frac{x}{x + 1} d x

f (x) = 3 x + 1, at x = 8

What is inverse oflaplace (0.9)/(0.1s^2+s+0.9) ?
The answer to inverse oflaplace (0.9)/(0.1s^2+s+0.9) is 9/4 e^{-5t}sinh(4t)