What is inverse oflaplace s/((s-1)^2+4) ?

The answer to inverse oflaplace s/((s-1)^2+4) is e^tcos(2t)+1/2 e^tsin(2t)

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inverse oflaplace s/((s-1)^2+4)

inverse laplace

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

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

Solution steps

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

Expand

\frac{s}{( s - 1 ) ^{2} + 4} : \frac{s - 1}{( s - 1 ) ^{2} + 4} + 1 \cdot \frac{1}{( s - 1 ) ^{2} + 4}

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

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

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

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

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

Refine

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

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

\frac{d}{d x} (5^{x} + 7^{x})

\frac{d}{d x} (- 262144 cos (2 x))

x \to - 3 lim (h (x))

x \to 3 lim (\frac{x ^{2}}{x - 3})

x \to 2 + lim (\frac{4}{x})

What is inverse oflaplace s/((s-1)^2+4) ?
The answer to inverse oflaplace s/((s-1)^2+4) is e^tcos(2t)+1/2 e^tsin(2t)