What is inverse oflaplace (s+7)/(s^2+8s+25) ?

The answer to inverse oflaplace (s+7)/(s^2+8s+25) is e^{-4t}cos(3t)+e^{-4t}sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (s+7)/(s^2+8s+25)

inverse laplace

\frac{s + 7}{s ^{2} + 8 s + 25}

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

Solution steps

L^{- 1} {\frac{s + 7}{s ^{2} + 8 s + 25}}

Expand

\frac{s + 7}{s ^{2} + 8 s + 25} : \frac{s + 4}{( s + 4 ) ^{2} + 9} + 3 \cdot \frac{1}{( s + 4 ) ^{2} + 9}

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

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

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

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

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

Refine

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

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

g (x) = (3 - 4 x)^{3}

\frac{d}{d x} (\frac{10 x}{x ^{2} + 4})

x \to 1 - lim (\frac{2 x}{ln ( x )})

\int (6 x + 5) d x

\int \frac{1}{( x + 3 ) ( x + 2 )} d x

What is inverse oflaplace (s+7)/(s^2+8s+25) ?
The answer to inverse oflaplace (s+7)/(s^2+8s+25) is e^{-4t}cos(3t)+e^{-4t}sin(3t)