What is inverse oflaplace s/(s^2-2s+26) ?

The answer to inverse oflaplace s/(s^2-2s+26) is e^tcos(5t)+1/5 e^tsin(5t)

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inverse oflaplace s/(s^2-2s+26)

inverse laplace

\frac{s}{s ^{2} - 2 s + 26}

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

Solution steps

L^{- 1} {\frac{s}{s ^{2} - 2 s + 26}}

Expand

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

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

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

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

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

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

Refine

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

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

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