What is inverse oflaplace (3s^2)/((s+2)(s+5)) ?

The answer to inverse oflaplace (3s^2)/((s+2)(s+5)) is 3\delta(t)+4e^{-2t}-25e^{-5t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (3s^2)/((s+2)(s+5))

inverse laplace

\frac{3 s ^{2}}{( s + 2 ) ( s + 5 )}

3 δ (t) + 4 e^{- 2 t} - 25 e^{- 5 t}

Solution steps

L^{- 1} {\frac{3 s ^{2}}{( s + 2 ) ( s + 5 )}}

Take the partial fraction of

\frac{3 s ^{2}}{( s + 2 ) ( s + 5 )} : 3 + \frac{4}{s + 2} - \frac{25}{s + 5}

= L^{- 1} {3 + \frac{4}{s + 2} - \frac{25}{s + 5}}

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} {3} + L^{- 1} {\frac{4}{s + 2}} - L^{- 1} {\frac{25}{s + 5}}

L^{- 1} {3} : 3 δ (t)

L^{- 1} {\frac{4}{s + 2}} : 4 e^{- 2 t}

L^{- 1} {\frac{25}{s + 5}} : 25 e^{- 5 t}

= 3 δ (t) + 4 e^{- 2 t} - 25 e^{- 5 t}

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

f (x) = (4 x + 3)^{2}

x \to 0 lim (tan^{x} (x))

2 x - 4

\int_{- 8}^{0} \frac{y}{8} + y + 9 d y

What is inverse oflaplace (3s^2)/((s+2)(s+5)) ?
The answer to inverse oflaplace (3s^2)/((s+2)(s+5)) is 3\delta(t)+4e^{-2t}-25e^{-5t}