What is inverse oflaplace s/((s-3)(s-4)(s-12)) ?

The answer to inverse oflaplace s/((s-3)(s-4)(s-12)) is 1/3 e^{3t}-1/2 e^{4t}+1/6 e^{12t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace s/((s-3)(s-4)(s-12))

inverse laplace

\frac{s}{( s - 3 ) ( s - 4 ) ( s - 12 )}

\frac{1}{3} e^{3 t} - \frac{1}{2} e^{4 t} + \frac{1}{6} e^{12 t}

Solution steps

L^{- 1} {\frac{s}{( s - 3 ) ( s - 4 ) ( s - 12 )}}

Take the partial fraction of

\frac{s}{( s - 3 ) ( s - 4 ) ( s - 12 )} : \frac{1}{3 ( s - 3 )} - \frac{1}{2 ( s - 4 )} + \frac{1}{6 ( s - 12 )}

= L^{- 1} {\frac{1}{3 ( s - 3 )} - \frac{1}{2 ( s - 4 )} + \frac{1}{6 ( s - 12 )}}

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}{3 ( s - 3 )}} - L^{- 1} {\frac{1}{2 ( s - 4 )}} + L^{- 1} {\frac{1}{6 ( s - 12 )}}

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

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

L^{- 1} {\frac{1}{6 ( s - 12 )}} : \frac{1}{6} e^{12 t}

= \frac{1}{3} e^{3 t} - \frac{1}{2} e^{4 t} + \frac{1}{6} e^{12 t}

\frac{\partial}{\partial x} (2 x^{\frac{3}{2}})

y = - x^{2} + 6 x, y = x^{2} - 2 x

\frac{3}{5 x ^{4}}

4 y^{^{''}} + 3 y^{^{'}} + 7 y = 0

x y^{^{'}} + y = x^{2} + 1

What is inverse oflaplace s/((s-3)(s-4)(s-12)) ?
The answer to inverse oflaplace s/((s-3)(s-4)(s-12)) is 1/3 e^{3t}-1/2 e^{4t}+1/6 e^{12t}