What is inverse oflaplace (2s)/((s-1)(s^2+25)) ?

The answer to inverse oflaplace (2s)/((s-1)(s^2+25)) is 1/13 e^t-1/13 cos(5t)+(5sin(5t))/(13)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (2s)/((s-1)(s^2+25))

inverse laplace

\frac{2 s}{( s - 1 ) ( s ^{2} + 25 )}

\frac{1}{13} e^{t} - \frac{1}{13} cos (5 t) + \frac{5 sin ( 5 t )}{13}

Solution steps

L^{- 1} {\frac{2 s}{( s - 1 ) ( s ^{2} + 25 )}}

Take the partial fraction of

\frac{2 s}{( s - 1 ) ( s ^{2} + 25 )} : \frac{1}{13 ( s - 1 )} + \frac{- s + 25}{13 ( s ^{2} + 25 )}

= L^{- 1} {\frac{1}{13 ( s - 1 )} + \frac{- s + 25}{13 ( s ^{2} + 25 )}}

Expand

= L^{- 1} {- \frac{s}{13 ( s ^{2} + 25 )} + \frac{1}{13 ( s - 1 )} + \frac{25}{13 ( s ^{2} + 25 )}}

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)}

= \frac{1}{13} L^{- 1} {\frac{1}{s - 1}} - \frac{1}{13} L^{- 1} {\frac{s}{s ^{2} + 25}} + \frac{25}{13} L^{- 1} {\frac{1}{s ^{2} + 25}}

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

L^{- 1} {\frac{s}{s ^{2} + 25}} : cos (5 t)

L^{- 1} {\frac{1}{s ^{2} + 25}} : \frac{1}{5} sin (5 t)

= \frac{1}{13} e^{t} - \frac{1}{13} cos (5 t) + \frac{25}{13} \cdot \frac{1}{5} sin (5 t)

Refine

\frac{1}{13} e^{t} - \frac{1}{13} cos (5 t) + \frac{25}{13} \frac{1}{5} sin (5 t) : \frac{1}{13} e^{t} - \frac{1}{13} cos (5 t) + \frac{5 sin ( 5 t )}{13}

= \frac{1}{13} e^{t} - \frac{1}{13} cos (5 t) + \frac{5 sin ( 5 t )}{13}

\frac{\partial}{\partial y} (e^{- y} (x^{2} + y^{2}))

x (\frac{d y}{d x}) = x^{2} - 4

f (x) = (5 x^{2} - 7 x) e^{x}

f (x) = ln^{3} (x)

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

What is inverse oflaplace (2s)/((s-1)(s^2+25)) ?
The answer to inverse oflaplace (2s)/((s-1)(s^2+25)) is 1/13 e^t-1/13 cos(5t)+(5sin(5t))/(13)