What is inverse oflaplace ((-2s+21))/((s^2-8s+20)) ?

The answer to inverse oflaplace ((-2s+21))/((s^2-8s+20)) is -2e^{4t}cos(2t)+13/2 e^{4t}sin(2t)

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inverse oflaplace ((-2s+21))/((s^2-8s+20))

inverse laplace

\frac{( - 2 s + 21 )}{( s ^{2} - 8 s + 20 )}

- 2 e^{4 t} cos (2 t) + \frac{13}{2} e^{4 t} sin (2 t)

Solution steps

L^{- 1} {\frac{( - 2 s + 21 )}{( s ^{2} - 8 s + 20 )}}

Expand

\frac{- 2 s + 21}{s ^{2} - 8 s + 20} : - 2 \cdot \frac{s - 4}{( s - 4 ) ^{2} + 4} + 13 \cdot \frac{1}{( s - 4 ) ^{2} + 4}

= L^{- 1} {- 2 \cdot \frac{s - 4}{( s - 4 ) ^{2} + 4} + 13 \cdot \frac{1}{( s - 4 ) ^{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)}

= - 2 L^{- 1} {\frac{s - 4}{( s - 4 ) ^{2} + 4}} + 13 L^{- 1} {\frac{1}{( s - 4 ) ^{2} + 4}}

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

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

= - 2 e^{4 t} cos (2 t) + 13 e^{4 t} \frac{1}{2} sin (2 t)

Refine

- 2 e^{4 t} cos (2 t) + 13 e^{4 t} \frac{1}{2} sin (2 t) : - 2 e^{4 t} cos (2 t) + \frac{13}{2} e^{4 t} sin (2 t)

= - 2 e^{4 t} cos (2 t) + \frac{13}{2} e^{4 t} sin (2 t)

\frac{d}{d x} (x (x^{3} + 7)^{4})

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2} - 4)

y (θ) = \frac{π}{3} sin (θ)

\frac{\partial}{\partial x} (9 x cos (3 x y))

\frac{d}{d x} ((x + a)^{2})

What is inverse oflaplace ((-2s+21))/((s^2-8s+20)) ?
The answer to inverse oflaplace ((-2s+21))/((s^2-8s+20)) is -2e^{4t}cos(2t)+13/2 e^{4t}sin(2t)