What is inverse oflaplace (104)/(s^2+8s+52) ?

The answer to inverse oflaplace (104)/(s^2+8s+52) is 52/3 e^{-4t}sin(6t)

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inverse oflaplace (104)/(s^2+8s+52)

inverse laplace

\frac{104}{s ^{2} + 8 s + 52}

\frac{52}{3} e^{- 4 t} sin (6 t)

Solution steps

L^{- 1} {\frac{104}{s ^{2} + 8 s + 52}}

Expand

\frac{104}{s ^{2} + 8 s + 52} : 104 \cdot \frac{1}{( s + 4 ) ^{2} + 36}

= L^{- 1} {104 \cdot \frac{1}{( s + 4 ) ^{2} + 36}}

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)}

= 104 L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 36}}

L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 36}} : e^{- 4 t} \frac{1}{6} sin (6 t)

= 104 e^{- 4 t} \frac{1}{6} sin (6 t)

Refine

104 e^{- 4 t} \frac{1}{6} sin (6 t) : \frac{52}{3} e^{- 4 t} sin (6 t)

= \frac{52}{3} e^{- 4 t} sin (6 t)

\frac{d}{d x} (\frac{x ^{3} - 7}{x ^{2} + 2})

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

\frac{d}{d x} (3 x^{2} + 2 x - 4)

\int \frac{x}{( 4 + x ^{2} )} d x

What is inverse oflaplace (104)/(s^2+8s+52) ?
The answer to inverse oflaplace (104)/(s^2+8s+52) is 52/3 e^{-4t}sin(6t)