What is inverse oflaplace 6/(s(s+2)^2(s+3)) ?

The answer to inverse oflaplace 6/(s(s+2)^2(s+3)) is 1/2 \H(t)+3/2 e^{-2t}-e^{-2t}*3t-2e^{-3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 6/(s(s+2)^2(s+3))

inverse laplace

\frac{6}{s ( s + 2 ) ^{2} ( s + 3 )}

\frac{1}{2} H (t) + \frac{3}{2} e^{- 2 t} - e^{- 2 t} \cdot 3 t - 2 e^{- 3 t}

Solution steps

L^{- 1} {\frac{6}{s ( s + 2 ) ^{2} ( s + 3 )}}

Take the partial fraction of

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

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

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

L^{- 1} {\frac{1}{2 s}} : \frac{1}{2} H (t)

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

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

L^{- 1} {\frac{2}{s + 3}} : 2 e^{- 3 t}

= \frac{1}{2} H (t) + \frac{3}{2} e^{- 2 t} - e^{- 2 t} \cdot 3 t - 2 e^{- 3 t}

\int \frac{1}{7 x ^{4}} d x

\int e^{x} \cdot sin (x - π) d x

x^{4} + 4 x^{2} - x

x \to - 1 lim (f (x) g (x))

\frac{\partial}{\partial x} (y^{3} e^{x y})

What is inverse oflaplace 6/(s(s+2)^2(s+3)) ?
The answer to inverse oflaplace 6/(s(s+2)^2(s+3)) is 1/2 \H(t)+3/2 e^{-2t}-e^{-2t}*3t-2e^{-3t}