What is inverse oflaplace (20)/(s^2+6s+144) ?

The answer to inverse oflaplace (20)/(s^2+6s+144) is (4sqrt(5))/(3sqrt(3))e^{-3t}sin(3sqrt(15)t)

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inverse oflaplace (20)/(s^2+6s+144)

inverse laplace

\frac{20}{s ^{2} + 6 s + 144}

\frac{4 5}{3 3} e^{- 3 t} sin (315 t)

Solution steps

L^{- 1} {\frac{20}{s ^{2} + 6 s + 144}}

Expand

\frac{20}{s ^{2} + 6 s + 144} : 20 \cdot \frac{1}{( s + 3 ) ^{2} + 135}

= L^{- 1} {20 \cdot \frac{1}{( s + 3 ) ^{2} + 135}}

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)}

= 20 L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 135}}

L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 135}} : e^{- 3 t} \frac{1}{3 15} sin (315 t)

= 20 e^{- 3 t} \frac{1}{3 15} sin (315 t)

Refine

20 e^{- 3 t} \frac{1}{3 15} sin (315 t) : \frac{4 5}{3 3} e^{- 3 t} sin (315 t)

= \frac{4 5}{3 3} e^{- 3 t} sin (315 t)

\int x^{4} - 2 x d x

f (x) = (2^{x})^{8}

x \to 0 - lim (\frac{1}{∣ x ∣})

\int z e^{z} d z

\frac{1}{( s - 1 ) ^{4}}

What is inverse oflaplace (20)/(s^2+6s+144) ?
The answer to inverse oflaplace (20)/(s^2+6s+144) is (4sqrt(5))/(3sqrt(3))e^{-3t}sin(3sqrt(15)t)