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

The answer to inverse oflaplace s/((s^2+25)(s-1)) is -1/26 cos(5t)+(5sin(5t))/(26)+1/26 e^t

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

inverse laplace

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

- \frac{1}{26} cos (5 t) + \frac{5 sin ( 5 t )}{26} + \frac{1}{26} e^{t}

Solution steps

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

Take the partial fraction of

\frac{s}{( s ^{2} + 25 ) ( s - 1 )} : \frac{- s + 25}{26 ( s ^{2} + 25 )} + \frac{1}{26 ( s - 1 )}

= L^{- 1} {\frac{- s + 25}{26 ( s ^{2} + 25 )} + \frac{1}{26 ( s - 1 )}}

Expand

= L^{- 1} {- \frac{s}{26 ( s ^{2} + 25 )} + \frac{25}{26 ( s ^{2} + 25 )} + \frac{1}{26 ( 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}{26} L^{- 1} {\frac{s}{s ^{2} + 25}} + \frac{25}{26} L^{- 1} {\frac{1}{s ^{2} + 25}} + \frac{1}{26} 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}{26} cos (5 t) + \frac{25}{26} \cdot \frac{1}{5} sin (5 t) + \frac{1}{26} e^{t}

Refine

- \frac{1}{26} cos (5 t) + \frac{25}{26} \frac{1}{5} sin (5 t) + \frac{1}{26} e^{t} : - \frac{1}{26} cos (5 t) + \frac{5 sin ( 5 t )}{26} + \frac{1}{26} e^{t}

= - \frac{1}{26} cos (5 t) + \frac{5 sin ( 5 t )}{26} + \frac{1}{26} e^{t}

x (x + 1) \frac{d y}{d x} + x y = 1, y (e) = 1

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

x \to 6 + lim (\frac{x}{x - 6})

\frac{1}{x ^{2}} 4

\frac{\partial}{\partial x} (\frac{6 x - 8 y}{6 x + 8 y})

What is inverse oflaplace s/((s^2+25)(s-1)) ?
The answer to inverse oflaplace s/((s^2+25)(s-1)) is -1/26 cos(5t)+(5sin(5t))/(26)+1/26 e^t