What is inverse oflaplace (2s+1)/(3s^2+5s) ?

The answer to inverse oflaplace (2s+1)/(3s^2+5s) is 1/5 \H(t)+7/15 e^{-(5t)/3}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (2s+1)/(3s^2+5s)

inverse laplace

\frac{2 s + 1}{3 s ^{2} + 5 s}

\frac{1}{5} H (t) + \frac{7}{15} e^{- \frac{5 t}{3}}

Solution steps

L^{- 1} {\frac{2 s + 1}{3 s ^{2} + 5 s}}

Take the partial fraction of

\frac{2 s + 1}{3 s ^{2} + 5 s} : \frac{1}{5 s} + \frac{7}{5 ( 3 s + 5 )}

= L^{- 1} {\frac{1}{5 s} + \frac{7}{5 ( 3 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{7}{5 ( 3 s + 5 )}}

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

L^{- 1} {\frac{7}{5 ( 3 s + 5 )}} : \frac{7}{15} e^{- \frac{5 t}{3}}

= \frac{1}{5} H (t) + \frac{7}{15} e^{- \frac{5 t}{3}}

\frac{\partial}{\partial y} (x sin (y) + y cos (x))

\int \frac{x ^{2} - 4}{x} d x

\frac{d}{d x} (x^{2} 2 x - 3)

x^{2} y^{^{'}} + \frac{4}{3} x y = - 4 cos (3 x) y^{- \frac{1}{2}}

What is inverse oflaplace (2s+1)/(3s^2+5s) ?
The answer to inverse oflaplace (2s+1)/(3s^2+5s) is 1/5 \H(t)+7/15 e^{-(5t)/3}