What is inverse oflaplace 1/(s^2+5s) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/(s^2+5s)

inverse laplace

\frac{1}{s ^{2} + 5 s}

\frac{1}{5} H (t) - \frac{1}{5} e^{- 5 t}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{5 s}} : \frac{1}{5} H (t)

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

= \frac{1}{5} H (t) - \frac{1}{5} e^{- 5 t}

y = x^{3} - 2 x^{2} + 5, (2, 5)

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 8

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

\int \frac{x + 3}{x ^{2} + 4} d x

\int \frac{1}{3 sin ( x ) + 4 cos ( x )} d x

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