What is inverse oflaplace (4-4s)/(4s^2+4s+17) ?

The answer to inverse oflaplace (4-4s)/(4s^2+4s+17) is -e^{-t/2}cos(2t)+3/4 e^{-t/2}sin(2t)

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inverse oflaplace (4-4s)/(4s^2+4s+17)

inverse laplace

\frac{4 - 4 s}{4 s ^{2} + 4 s + 17}

- e^{- \frac{t}{2}} cos (2 t) + \frac{3}{4} e^{- \frac{t}{2}} sin (2 t)

Solution steps

L^{- 1} {\frac{4 - 4 s}{4 s ^{2} + 4 s + 17}}

Expand

\frac{4 - 4 s}{4 s ^{2} + 4 s + 17} : - \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + 4} + \frac{3}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + 4}

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

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

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

= - e^{- \frac{t}{2}} cos (2 t) + \frac{3}{2} e^{- \frac{t}{2}} \frac{1}{2} sin (2 t)

Refine

- e^{- \frac{t}{2}} cos (2 t) + \frac{3}{2} e^{- \frac{t}{2}} \frac{1}{2} sin (2 t) : - e^{- \frac{t}{2}} cos (2 t) + \frac{3}{4} e^{- \frac{t}{2}} sin (2 t)

= - e^{- \frac{t}{2}} cos (2 t) + \frac{3}{4} e^{- \frac{t}{2}} sin (2 t)

f (x) = sec (x) + tan (x)

\int 8 x - 4 d x

\int 4 sin^{3} (6 x co) s^{3} 6 x d x

\frac{d ^{2} y}{d t ^{2}} - 9 y = 0

\int (π^{3} - 1) d x

What is inverse oflaplace (4-4s)/(4s^2+4s+17) ?
The answer to inverse oflaplace (4-4s)/(4s^2+4s+17) is -e^{-t/2}cos(2t)+3/4 e^{-t/2}sin(2t)