What is inverse oflaplace ((2s+13))/(s^2+4s+13) ?

The answer to inverse oflaplace ((2s+13))/(s^2+4s+13) is 2e^{-2t}cos(3t)+3e^{-2t}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+13))/(s^2+4s+13)

inverse laplace

\frac{( 2 s + 13 )}{s ^{2} + 4 s + 13}

2 e^{- 2 t} cos (3 t) + 3 e^{- 2 t} sin (3 t)

Solution steps

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

Expand

\frac{2 s + 13}{s ^{2} + 4 s + 13} : 2 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 9} + 9 \cdot \frac{1}{( s + 2 ) ^{2} + 9}

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

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

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

= 2 e^{- 2 t} cos (3 t) + 9 e^{- 2 t} \frac{1}{3} sin (3 t)

Refine

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

= 2 e^{- 2 t} cos (3 t) + 3 e^{- 2 t} sin (3 t)

\int_{0}^{4} 1

\frac{d}{d x} (7 x^{- 1})

\int 2 x sin (3 x y) d x

x \to - 4 lim (x^{3} + x^{2})

\int_{2}^{5} f (x) d x

What is inverse oflaplace ((2s+13))/(s^2+4s+13) ?
The answer to inverse oflaplace ((2s+13))/(s^2+4s+13) is 2e^{-2t}cos(3t)+3e^{-2t}sin(3t)