What is inverse oflaplace (2s-3)/(2s^2-8s+16) ?

The answer to inverse oflaplace (2s-3)/(2s^2-8s+16) is e^{2t}cos(2t)+1/4 e^{2t}sin(2t)

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inverse oflaplace (2s-3)/(2s^2-8s+16)

inverse laplace

\frac{2 s - 3}{2 s ^{2} - 8 s + 16}

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

Solution steps

L^{- 1} {\frac{2 s - 3}{2 s ^{2} - 8 s + 16}}

Expand

\frac{2 s - 3}{2 s ^{2} - 8 s + 16} : \frac{s - 2}{( s - 2 ) ^{2} + 4} + \frac{1}{2} \cdot \frac{1}{( s - 2 ) ^{2} + 4}

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

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

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

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

Refine

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

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

f (x) = 4 x^{3} ln (x)

x \to 1 lim (tan (\frac{x}{2}))

y^{^{''}} - y = e^{t} (t + e^{t})

2 t (3 t^{3} - 7)^{4}

x \to \infty lim (\frac{ln ( x )}{e ^{x}})

What is inverse oflaplace (2s-3)/(2s^2-8s+16) ?
The answer to inverse oflaplace (2s-3)/(2s^2-8s+16) is e^{2t}cos(2t)+1/4 e^{2t}sin(2t)