What is inverse oflaplace (s-4)/(s^2-8s+10) ?

The answer to inverse oflaplace (s-4)/(s^2-8s+10) is e^{4t}cosh(sqrt(6)t)

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inverse oflaplace (s-4)/(s^2-8s+10)

inverse laplace

\frac{s - 4}{s ^{2} - 8 s + 10}

e^{4 t} cosh (6 t)

Solution steps

L^{- 1} {\frac{s - 4}{s ^{2} - 8 s + 10}}

\frac{s - 4}{s ^{2} - 8 s + 10} = \frac{s - 4}{( s - 4 ) ^{2} - 6}

= L^{- 1} {\frac{s - 4}{( s - 4 ) ^{2} - 6}}

Apply inverse transform rule:

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

then

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

For

\frac{s - 4}{( s - 4 ) ^{2} - 6} : a = 4, F (s) = \frac{s}{s ^{2} - 6}

= e^{4 t} L^{- 1} {\frac{s}{s ^{2} - 6}}

L^{- 1} {\frac{s}{s ^{2} - 6}} : cosh (6 t)

= e^{4 t} cosh (6 t)

t^{\frac{4}{3}}

f (x) = x^{3} - 9 x + 4, (3, 4)

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

(1 + x)^{- 5}

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

What is inverse oflaplace (s-4)/(s^2-8s+10) ?
The answer to inverse oflaplace (s-4)/(s^2-8s+10) is e^{4t}cosh(sqrt(6)t)