What is inverse oflaplace ((3s-4))/(s^2+9) ?

The answer to inverse oflaplace ((3s-4))/(s^2+9) is 3cos(3t)-4/3 sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace ((3s-4))/(s^2+9)

inverse laplace

\frac{( 3 s - 4 )}{s ^{2} + 9}

3 cos (3 t) - \frac{4}{3} sin (3 t)

Solution steps

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

Expand

= L^{- 1} {\frac{3 s}{s ^{2} + 9} - \frac{4}{s ^{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)}

= 3 L^{- 1} {\frac{s}{s ^{2} + 9}} - 4 L^{- 1} {\frac{1}{s ^{2} + 9}}

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

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

= 3 cos (3 t) - 4 \cdot \frac{1}{3} sin (3 t)

Refine

3 cos (3 t) - 4 \frac{1}{3} sin (3 t) : 3 cos (3 t) - \frac{4}{3} sin (3 t)

= 3 cos (3 t) - \frac{4}{3} sin (3 t)

\int_{0}^{9} 2 y d y

\frac{d}{d θ} (sec (θ) tan (θ))

\frac{3}{s - 9}

\frac{d}{d x} (f (x) g (x))

\frac{y}{y + \frac{a}{y}}

What is inverse oflaplace ((3s-4))/(s^2+9) ?
The answer to inverse oflaplace ((3s-4))/(s^2+9) is 3cos(3t)-4/3 sin(3t)