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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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

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

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

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

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

Refine

- \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{1}{13} e^{t}

y = \frac{2 x}{x ^{2} - 1}

y (θ) = e^{t a n (θ)}

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

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

\frac{6 x}{x ^{2} + 2}

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