What is inverse oflaplace (2/3)/(s^2+2s+17) ?

The answer to inverse oflaplace (2/3)/(s^2+2s+17) is (e^{-t}sin(4t))/6

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inverse oflaplace (2/3)/(s^2+2s+17)

inverse laplace

\frac{\frac{2}{3}}{s ^{2} + 2 s + 17}

\frac{e ^{- t} sin ( 4 t )}{6}

Solution steps

L^{- 1} {\frac{\frac{2}{3}}{s ^{2} + 2 s + 17}}

Expand

\frac{\frac{2}{3}}{s ^{2} + 2 s + 17} : \frac{2}{3} \cdot \frac{1}{( s + 1 ) ^{2} + 16}

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

= \frac{2}{3} L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 16}}

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

= \frac{2}{3} e^{- t} \frac{1}{4} sin (4 t)

Refine

\frac{2}{3} e^{- t} \frac{1}{4} sin (4 t) : \frac{e ^{- t} sin ( 4 t )}{6}

= \frac{e ^{- t} sin ( 4 t )}{6}

\frac{25 - x ^{2}}{32 x}

\int (tan (5 x) - sec (5 x))^{2} d x

\frac{d}{d x} (2^{x})

\frac{d y}{d x} = y^{2} e^{x} sin (3 x)

\frac{d}{d x} (\frac{2}{x ^{3}} - \frac{1}{x ^{2}})

What is inverse oflaplace (2/3)/(s^2+2s+17) ?
The answer to inverse oflaplace (2/3)/(s^2+2s+17) is (e^{-t}sin(4t))/6