inverse oflaplace f(s)= 1/(s(s-3))

⇥1) You can decompose into partial fractions:

⇥

\frac{1}{s \cdot (s - 3)} = \frac{A}{s} + \frac{B}{s - 3}

⇥For the simple factor:

s

and its root:

0

, compute A by multiplying by the denominator:

s \cdot (s - 3)

and evaluating at the root (the other terms cancel out):

⇥2) Let's solve in

ℝ

(0 - 3) \cdot A = 1

⇥

- 3 A = 1

⇥

A = \frac{1}{-3}

⇥Solution found:

A = - \frac{1}{3}

⇥For the simple factor:

s - 3

and its root:

3

, compute B by multiplying by the denominator:

s \cdot (s - 3)

and evaluating at the root (the other terms cancel out):

⇥3) Let's solve in

ℝ

3 B = 1

⇥Solution obtained:

B = \frac{1}{3}

⇥4) Finally, substitute the unknowns back into the initial partial fraction decomposition:

⇥

- \frac{\frac{1}{3}}{s} + \frac{\frac{1}{3}}{s - 3}

⇥

- \frac{1}{3} \cdot \frac{1}{s} + \frac{1}{3} \cdot \frac{1}{s - 3}

⇥ℹ

- \frac{1}{3} \cdot \frac{1}{s}

- \frac{1}{3 s}

‖

\frac{1}{3} \cdot \frac{1}{s - 3}

\frac{1}{3 \cdot (s - 3)}

⇥

- \frac{1}{3 s} + \frac{1}{3 \cdot (s - 3)}

Find the causal inverse transform, for t ≥ 0. By linearity, invert each partial fraction

The formula L⁻¹(1/(s-a)^n) = t^(n-1)e^(at)/(n-1)! gives for

\frac{1}{3 \cdot (s - 3)}

\frac{\exp (3 t)}{3 \cdot 0!}

The formula L⁻¹(1/(s-a)^n) = t^(n-1)e^(at)/(n-1)! gives for

- \frac{1}{3 s}

\frac{-1 \cdot \exp (0 \cdot t)}{3 \cdot 0!}

⇥

\frac{\exp (3 t)}{3 \cdot 0!} + \frac{-1 \cdot \exp (0 \cdot t)}{3 \cdot 0!}

⇥ℹ

3 \cdot 1

3

‖

0 \cdot t

0

‖

\exp (0)

1

▶Inverse Laplace transform:

\frac{\exp (3 t)}{3} - \frac{1}{3}