What is inverse oflaplace (s-1)/(s^2-2s+5) ?

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

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inverse oflaplace (s-1)/(s^2-2s+5)

inverse laplace

\frac{s - 1}{s ^{2} - 2 s + 5}

e^{t} cos (2 t)

Solution steps

L^{- 1} {\frac{s - 1}{s ^{2} - 2 s + 5}}

\frac{s - 1}{s ^{2} - 2 s + 5} = \frac{s - 1}{( s - 1 ) ^{2} + 4}

= L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 4}}

Apply inverse transform rule:

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

then

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

For

\frac{s - 1}{( s - 1 ) ^{2} + 4} : a = 1, F (s) = \frac{s}{s ^{2} + 4}

= e^{t} L^{- 1} {\frac{s}{s ^{2} + 4}}

L^{- 1} {\frac{s}{s ^{2} + 4}} : cos (2 t)

= e^{t} cos (2 t)

\int \frac{4}{t} d t

t an g e n t y = x^{3} - 6 x^{2} + 7 x + 3

\int \frac{( e ^{x} - e ^{- x} )}{e ^{x} + e ^{- x}} d x

\frac{x d y}{d x} = 5 x e^{x} - y + 9 x^{2}

\int 5 x^{2} e^{6 x} d x

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