What is inverse oflaplace (2s)/(s^2+8s+25) ?

The answer to inverse oflaplace (2s)/(s^2+8s+25) is 2e^{-4t}cos(3t)-8/3 e^{-4t}sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (2s)/(s^2+8s+25)

inverse laplace

\frac{2 s}{s ^{2} + 8 s + 25}

2 e^{- 4 t} cos (3 t) - \frac{8}{3} e^{- 4 t} sin (3 t)

Solution steps

L^{- 1} {\frac{2 s}{s ^{2} + 8 s + 25}}

Expand

\frac{2 s}{s ^{2} + 8 s + 25} : 2 \cdot \frac{s + 4}{( s + 4 ) ^{2} + 9} - 8 \cdot \frac{1}{( s + 4 ) ^{2} + 9}

= L^{- 1} {2 \cdot \frac{s + 4}{( s + 4 ) ^{2} + 9} - 8 \cdot \frac{1}{( s + 4 ) ^{2} + 9}}

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} + 9}} - 8 L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 9}}

L^{- 1} {\frac{s + 4}{( s + 4 ) ^{2} + 9}} : e^{- 4 t} cos (3 t)

L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 9}} : e^{- 4 t} \frac{1}{3} sin (3 t)

= 2 e^{- 4 t} cos (3 t) - 8 e^{- 4 t} \frac{1}{3} sin (3 t)

Refine

2 e^{- 4 t} cos (3 t) - 8 e^{- 4 t} \frac{1}{3} sin (3 t) : 2 e^{- 4 t} cos (3 t) - \frac{8}{3} e^{- 4 t} sin (3 t)

= 2 e^{- 4 t} cos (3 t) - \frac{8}{3} e^{- 4 t} sin (3 t)

\frac{d x}{d t} - 5 x = 2 t

10 x

\frac{d x}{d y}, x y = 1

\int x^{2} + 1 + \frac{1}{x ^{2} + 1} d x

\int (\frac{9}{x ^{2} + 9})^{2} d x

What is inverse oflaplace (2s)/(s^2+8s+25) ?
The answer to inverse oflaplace (2s)/(s^2+8s+25) is 2e^{-4t}cos(3t)-8/3 e^{-4t}sin(3t)