What is inverse oflaplace s/(s^2+6s+25) ?

The answer to inverse oflaplace s/(s^2+6s+25) is e^{-3t}cos(4t)-3/4 e^{-3t}sin(4t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace s/(s^2+6s+25)

inverse laplace

\frac{s}{s ^{2} + 6 s + 25}

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

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 6 s + 25}}

Expand

\frac{s}{s ^{2} + 6 s + 25} : \frac{s + 3}{( s + 3 ) ^{2} + 16} - 3 \cdot \frac{1}{( s + 3 ) ^{2} + 16}

= L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 16} - 3 \cdot \frac{1}{( s + 3 ) ^{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{s + 3}{( s + 3 ) ^{2} + 16}} - 3 L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 16}}

L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 16}} : e^{- 3 t} cos (4 t)

L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 16}} : e^{- 3 t} \frac{1}{4} sin (4 t)

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

Refine

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

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

\frac{d}{d x} (\frac{3 - 2 x - x ^{2}}{x ^{2} - 6})

\frac{d}{d x} (x^{4} + 4 x^{2})

\int - 3 cos (3 x) d x

\int 5^{2 x} d x

\frac{d}{d x} ((ln (2 x))^{2})

What is inverse oflaplace s/(s^2+6s+25) ?
The answer to inverse oflaplace s/(s^2+6s+25) is e^{-3t}cos(4t)-3/4 e^{-3t}sin(4t)