What is inverse oflaplace ((s+1))/(6s^2+7s+2) ?

The answer to inverse oflaplace ((s+1))/(6s^2+7s+2) is 1/2 e^{-t/2}-1/3 e^{-(2t)/3}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace ((s+1))/(6s^2+7s+2)

inverse laplace

\frac{( s + 1 )}{6 s ^{2} + 7 s + 2}

\frac{1}{2} e^{- \frac{t}{2}} - \frac{1}{3} e^{- \frac{2 t}{3}}

Solution steps

L^{- 1} {\frac{( s + 1 )}{6 s ^{2} + 7 s + 2}}

Take the partial fraction of

\frac{s + 1}{6 s ^{2} + 7 s + 2} : \frac{1}{2 s + 1} - \frac{1}{3 s + 2}

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

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

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

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

= \frac{1}{2} e^{- \frac{t}{2}} - \frac{1}{3} e^{- \frac{2 t}{3}}

x \to 2 lim (1 + 2 x e^{3 x})

x \to 0 - lim (∣ sin (x) ∣^{x})

\int \frac{1}{( x - 2 ) ^{2}} d x

\int 3 x^{2} d x

f (t) = t

What is inverse oflaplace ((s+1))/(6s^2+7s+2) ?
The answer to inverse oflaplace ((s+1))/(6s^2+7s+2) is 1/2 e^{-t/2}-1/3 e^{-(2t)/3}