What is inverse oflaplace ((s-2))/((s-2)^2-3) ?

The answer to inverse oflaplace ((s-2))/((s-2)^2-3) is e^{2t}cosh(sqrt(3)t)

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inverse oflaplace ((s-2))/((s-2)^2-3)

inverse laplace

\frac{( s - 2 )}{( s - 2 ) ^{2} - 3}

e^{2 t} cosh (3 t)

Solution steps

L^{- 1} {\frac{( s - 2 )}{( s - 2 ) ^{2} - 3}}

Apply inverse transform rule:

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

then

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

For

\frac{s - 2}{( s - 2 ) ^{2} - 3} : a = 2, F (s) = \frac{s}{s ^{2} - 3}

= e^{2 t} L^{- 1} {\frac{s}{s ^{2} - 3}}

L^{- 1} {\frac{s}{s ^{2} - 3}} : cosh (3 t)

= e^{2 t} cosh (3 t)

\int \frac{( e ^{x} )}{1 + e ^{x}} d x

y = 4 t t + 7

\int_{- 2}^{2} x^{2} - 1 d x

f (x) = 2 e^{- x}

\int 4 sin^{4} (x) cos^{2} (x) d x

What is inverse oflaplace ((s-2))/((s-2)^2-3) ?
The answer to inverse oflaplace ((s-2))/((s-2)^2-3) is e^{2t}cosh(sqrt(3)t)