What is inverse oflaplace ((s-3))/(s^2-4s+3) ?

The answer to inverse oflaplace ((s-3))/(s^2-4s+3) is e^{2t}cosh(t)-e^{2t}sinh(t)

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inverse oflaplace ((s-3))/(s^2-4s+3)

inverse laplace

\frac{( s - 3 )}{s ^{2} - 4 s + 3}

e^{2 t} cosh (t) - e^{2 t} sinh (t)

Solution steps

L^{- 1} {\frac{( s - 3 )}{s ^{2} - 4 s + 3}}

Expand

\frac{s - 3}{s ^{2} - 4 s + 3} : \frac{s - 2}{( s - 2 ) ^{2} - 1} - 1 \cdot \frac{1}{( s - 2 ) ^{2} - 1}

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

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

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

L^{- 1} {\frac{1}{( s - 2 ) ^{2} - 1}} : e^{2 t} sinh (t)

= e^{2 t} cosh (t) - 1 \cdot e^{2 t} sinh (t)

Refine

e^{2 t} cosh (t) - 1 e^{2 t} sinh (t) : e^{2 t} cosh (t) - e^{2 t} sinh (t)

= e^{2 t} cosh (t) - e^{2 t} sinh (t)

4.8 t^{\frac{1}{3}}

\int_{0}^{1} x e^{9 x} d x

\frac{d}{d x} ((a^{\frac{2}{3}} - x^{\frac{2}{3}})^{\frac{3}{2}})

\frac{\partial}{\partial y} (e^{- 9 t} cos (π x))

\frac{d}{d x} ((1 - sin (x))^{3})

What is inverse oflaplace ((s-3))/(s^2-4s+3) ?
The answer to inverse oflaplace ((s-3))/(s^2-4s+3) is e^{2t}cosh(t)-e^{2t}sinh(t)