What is inverse oflaplace s/((s-2)^2+16) ?

The answer to inverse oflaplace s/((s-2)^2+16) is e^{2t}cos(4t)+1/2 e^{2t}sin(4t)

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inverse oflaplace s/((s-2)^2+16)

inverse laplace

\frac{s}{( s - 2 ) ^{2} + 16}

e^{2 t} cos (4 t) + \frac{1}{2} e^{2 t} sin (4 t)

Solution steps

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

Expand

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

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

= e^{2 t} cos (4 t) + 2 e^{2 t} \frac{1}{4} sin (4 t)

Refine

e^{2 t} cos (4 t) + 2 e^{2 t} \frac{1}{4} sin (4 t) : e^{2 t} cos (4 t) + \frac{1}{2} e^{2 t} sin (4 t)

= e^{2 t} cos (4 t) + \frac{1}{2} e^{2 t} sin (4 t)

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

\int_{\frac{1}{3}}^{5} 16 x ln (3 x) d x

f (x) = x^{2} + 4 x - 5, at x = 2

\frac{d}{d x} (n ln (x))

\int \frac{1}{25 - 81 x ^{2}} d x

What is inverse oflaplace s/((s-2)^2+16) ?
The answer to inverse oflaplace s/((s-2)^2+16) is e^{2t}cos(4t)+1/2 e^{2t}sin(4t)