What is inverse oflaplace (2s+3)/(s^2-4s+20) ?

The answer to inverse oflaplace (2s+3)/(s^2-4s+20) is 2e^{2t}cos(4t)+7/4 e^{2t}sin(4t)

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inverse oflaplace (2s+3)/(s^2-4s+20)

inverse laplace

\frac{2 s + 3}{s ^{2} - 4 s + 20}

2 e^{2 t} cos (4 t) + \frac{7}{4} e^{2 t} sin (4 t)

Solution steps

L^{- 1} {\frac{2 s + 3}{s ^{2} - 4 s + 20}}

Expand

\frac{2 s + 3}{s ^{2} - 4 s + 20} : 2 \cdot \frac{s - 2}{( s - 2 ) ^{2} + 16} + 7 \cdot \frac{1}{( s - 2 ) ^{2} + 16}

= L^{- 1} {2 \cdot \frac{s - 2}{( s - 2 ) ^{2} + 16} + 7 \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)}

= 2 L^{- 1} {\frac{s - 2}{( s - 2 ) ^{2} + 16}} + 7 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)

= 2 e^{2 t} cos (4 t) + 7 e^{2 t} \frac{1}{4} sin (4 t)

Refine

2 e^{2 t} cos (4 t) + 7 e^{2 t} \frac{1}{4} sin (4 t) : 2 e^{2 t} cos (4 t) + \frac{7}{4} e^{2 t} sin (4 t)

= 2 e^{2 t} cos (4 t) + \frac{7}{4} e^{2 t} sin (4 t)

f (x) = x^{2} (1 - 2 x)

\frac{d y}{d x} = \frac{5 x}{y}

\frac{d}{d x} (\frac{x ^{4} - x + 1}{x ^{4} + x + 1})

\int cos (x^{3} - 1) d x

What is inverse oflaplace (2s+3)/(s^2-4s+20) ?
The answer to inverse oflaplace (2s+3)/(s^2-4s+20) is 2e^{2t}cos(4t)+7/4 e^{2t}sin(4t)