What is inverse oflaplace 2/(4(s-3)^2+64) ?

The answer to inverse oflaplace 2/(4(s-3)^2+64) is e^{3t} 1/8 sin(4t)

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inverse oflaplace 2/(4(s-3)^2+64)

inverse laplace

\frac{2}{4 ( s - 3 ) ^{2} + 64}

e^{3 t} \frac{1}{8} sin (4 t)

Solution steps

L^{- 1} {\frac{2}{4 ( s - 3 ) ^{2} + 64}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{2}{4 ( s - 3 ) ^{2} + 64} : a = 3, F (s) = \frac{2}{4 s ^{2} + 64}

= e^{3 t} L^{- 1} {\frac{2}{4 s ^{2} + 64}}

L^{- 1} {\frac{2}{4 s ^{2} + 64}} : \frac{1}{8} sin (4 t)

= e^{3 t} \frac{1}{8} sin (4 t)

\int cos^{4} (5 x + 1) d x

w \to - k lim (\frac{w ^{2} + 9 k w + 8 k ^{2}}{w ^{2} + k w})

\int arctan (2 x) d x

\int e^{- x} \cdot (2 - x) d x

What is inverse oflaplace 2/(4(s-3)^2+64) ?
The answer to inverse oflaplace 2/(4(s-3)^2+64) is e^{3t} 1/8 sin(4t)