What is inverse oflaplace 1/(s^2-2s+26) ?

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

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inverse oflaplace 1/(s^2-2s+26)

inverse laplace

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

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

Solution steps

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

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

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s - 1 ) ^{2} + 25} : a = 1, F (s) = \frac{1}{s ^{2} + 25}

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

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

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

\frac{\partial}{\partial y} (ln (1 + x^{2} + e^{y} + z))

\frac{s ^{2} + 6 s + 2}{s ( s + 1 ) ^{2}}

y^{^{'}} = t y - 2 t + y - 2

3 t^{3} - 2 t^{2} - 7 t + 2

\frac{\partial}{\partial x} (1^{x})

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