What is inverse oflaplace ((10s+3))/(s^2+4s+29) ?

The answer to inverse oflaplace ((10s+3))/(s^2+4s+29) is 10e^{-2t}cos(5t)-17/5 e^{-2t}sin(5t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace ((10s+3))/(s^2+4s+29)

inverse laplace

\frac{( 10 s + 3 )}{s ^{2} + 4 s + 29}

10 e^{- 2 t} cos (5 t) - \frac{17}{5} e^{- 2 t} sin (5 t)

Solution steps

L^{- 1} {\frac{( 10 s + 3 )}{s ^{2} + 4 s + 29}}

Expand

\frac{10 s + 3}{s ^{2} + 4 s + 29} : 10 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 25} - 17 \cdot \frac{1}{( s + 2 ) ^{2} + 25}

= L^{- 1} {10 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 25} - 17 \cdot \frac{1}{( s + 2 ) ^{2} + 25}}

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)}

= 10 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 25}} - 17 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 25}}

L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 25}} : e^{- 2 t} cos (5 t)

L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 25}} : e^{- 2 t} \frac{1}{5} sin (5 t)

= 10 e^{- 2 t} cos (5 t) - 17 e^{- 2 t} \frac{1}{5} sin (5 t)

Refine

10 e^{- 2 t} cos (5 t) - 17 e^{- 2 t} \frac{1}{5} sin (5 t) : 10 e^{- 2 t} cos (5 t) - \frac{17}{5} e^{- 2 t} sin (5 t)

= 10 e^{- 2 t} cos (5 t) - \frac{17}{5} e^{- 2 t} sin (5 t)

x \to 0 + lim (x^{- 5})

\frac{\partial}{\partial x} (8 x)

\int 4 π x^{2} d x

y = 8 x e^{x}, (0, 0)

f (x) = \frac{x ^{2} - 2 e ^{x}}{x + 3 e ^{x}}

What is inverse oflaplace ((10s+3))/(s^2+4s+29) ?
The answer to inverse oflaplace ((10s+3))/(s^2+4s+29) is 10e^{-2t}cos(5t)-17/5 e^{-2t}sin(5t)