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

The answer to inverse oflaplace ((s+6))/(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+6))/(s^2+6s+25)

inverse laplace

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

Expand

\frac{s + 6}{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)

0.001 x^{3} + 4 x + 1458

f (x) = \frac{1}{x ^{17}}

5 x^{^{''}} + 25 x^{^{'}} + 340 x = 0

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

2 x^{2} - 8 x - 55, - x^{2} - 2 x + 50, - 5, 7

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