What is inverse oflaplace (6s-4)/(s^2-4s+20) ?

The answer to inverse oflaplace (6s-4)/(s^2-4s+20) is 6e^{2t}cos(4t)+2e^{2t}sin(4t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace (6s-4)/(s^2-4s+20)

inverse laplace

\frac{6 s - 4}{s ^{2} - 4 s + 20}

6 e^{2 t} cos (4 t) + 2 e^{2 t} sin (4 t)

Solution steps

L^{- 1} {\frac{6 s - 4}{s ^{2} - 4 s + 20}}

Expand

\frac{6 s - 4}{s ^{2} - 4 s + 20} : 6 \cdot \frac{s - 2}{( s - 2 ) ^{2} + 16} + 8 \cdot \frac{1}{( s - 2 ) ^{2} + 16}

= L^{- 1} {6 \cdot \frac{s - 2}{( s - 2 ) ^{2} + 16} + 8 \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)}

= 6 L^{- 1} {\frac{s - 2}{( s - 2 ) ^{2} + 16}} + 8 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)

= 6 e^{2 t} cos (4 t) + 8 e^{2 t} \frac{1}{4} sin (4 t)

Refine

6 e^{2 t} cos (4 t) + 8 e^{2 t} \frac{1}{4} sin (4 t) : 6 e^{2 t} cos (4 t) + 2 e^{2 t} sin (4 t)

= 6 e^{2 t} cos (4 t) + 2 e^{2 t} sin (4 t)

What is inverse oflaplace (6s-4)/(s^2-4s+20) ?
The answer to inverse oflaplace (6s-4)/(s^2-4s+20) is 6e^{2t}cos(4t)+2e^{2t}sin(4t)