What is inverse oflaplace 4/s (1/(6+5s)) ?

The answer to inverse oflaplace 4/s (1/(6+5s)) is 2/3 \H(t)-2/3 e^{-(6t)/5}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 4/s (1/(6+5s))

inverse laplace

\frac{4}{s} (\frac{1}{6 + 5 s})

\frac{2}{3} H (t) - \frac{2}{3} e^{- \frac{6 t}{5}}

Solution steps

L^{- 1} {\frac{4}{s} (\frac{1}{5 s + 6})}

Expand

\frac{4}{s} \frac{1}{6 + 5 s} : \frac{4}{6 s + 5 s ^{2}}

= L^{- 1} {\frac{4}{6 s + 5 s ^{2}}}

Take the partial fraction of

\frac{4}{6 s + 5 s ^{2}} : \frac{2}{3 s} - \frac{10}{3 ( 5 s + 6 )}

= L^{- 1} {\frac{2}{3 s} - \frac{10}{3 ( 5 s + 6 )}}

Expand

= L^{- 1} {\frac{2}{3 s} - \frac{10}{15 s + 18}}

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{2}{3 s}} - L^{- 1} {\frac{10}{15 s + 18}}

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

L^{- 1} {\frac{10}{15 s + 18}} : \frac{2}{3} e^{- \frac{6 t}{5}}

= \frac{2}{3} H (t) - \frac{2}{3} e^{- \frac{6 t}{5}}

\int_{0}^{\frac{π}{2}} - 8 cos^{5} (x) d x

\frac{\partial}{\partial x} (3 x^{2} (5 x + 7 y))

\int 5 tan (x) sec^{4} (x) d x

tan (x) \frac{d y}{d x} = y

x \to 5 lim (\frac{x ^{2} - 3 x - 10}{x ^{2} - 10 x + 25})

What is inverse oflaplace 4/s (1/(6+5s)) ?
The answer to inverse oflaplace 4/s (1/(6+5s)) is 2/3 \H(t)-2/3 e^{-(6t)/5}