What is inverse oflaplace 6/s+(8(4))/(s^2+16) ?

The answer to inverse oflaplace 6/s+(8(4))/(s^2+16) is 6\H(t)+8sin(4t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 6/s+(8(4))/(s^2+16)

inverse laplace

\frac{6}{s} + \frac{8 ( 4 )}{s ^{2} + 16}

6 H (t) + 8 sin (4 t)

Solution steps

L^{- 1} {\frac{6}{s} + \frac{8 ( 4 )}{s ^{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)}

= L^{- 1} {\frac{6}{s}} + L^{- 1} {\frac{8 \cdot 4}{s ^{2} + 16}}

L^{- 1} {\frac{6}{s}} : 6 H (t)

L^{- 1} {\frac{8 \cdot 4}{s ^{2} + 16}} : 8 sin (4 t)

= 6 H (t) + 8 sin (4 t)

e^{11}

f (x) = x - 6 3 x

\int 2 - 32 x^{3} d x

x^{2} - 5 x + 2

x \to 1 - lim (x^{2} + 2)

What is inverse oflaplace 6/s+(8(4))/(s^2+16) ?
The answer to inverse oflaplace 6/s+(8(4))/(s^2+16) is 6\H(t)+8sin(4t)