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)

x \to 0 lim (\frac{ln ( x ^{2} + 1 )}{x})

x^{2}, x - 4, - 4, 4

f (x) = \frac{1}{10} x - \frac{1}{6}

f (x) = e^{\frac{x}{2}}

\int 5 t^{4} cos (t^{5} + 1) d t

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)