inverse laplace 1/(s^2-s) — step-by-step solution

⇥1) You can perform a partial fraction decomposition:

⇥

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

⇥For the simple factor:

s

and its root:

0

, the unknown A can be easily computed by multiplying by the denominator:

s \cdot (s - 1)

and evaluating at the root because the other terms cancel out:

⇥2) Let's solve in

ℝ

(0 - 1) \cdot A = 1

⇥ⓘ Adding 0 changes nothing:

0 - 1

-1

■Multiplying by -1 gives the opposite:

-1 \cdot A

- A

⇥

- A = 1

⇥ⓘ The sign of both sides is changed

⇥Solution obtained:

A = -1

⇥For the simple factor:

s - 1

and its root:

1

, compute B by multiplying by the denominator:

s \cdot (s - 1)

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

⇥3) Let's solve in

ℝ

1 \cdot B = 1

⇥Solution found:

B = 1

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

⇥

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

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

The inverse Laplace formula for

\frac{1}{{(s - a)}^{n}}

t^{n - 1} \cdot \frac{\exp (a t)}{(n - 1)!}

; it gives for

\frac{1}{s - 1}

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

The inverse Laplace formula for

\frac{1}{{(s - a)}^{n}}

t^{n - 1} \cdot \frac{\exp (a t)}{(n - 1)!}

; it gives for

- \frac{1}{s}

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

⇥

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

⇥ⓘ By convention, the factorial of 0 is 1:

0!

1

■A product with a zero factor is zero:

0 \cdot t

0

■Any nonzero number raised to the power 0 is 1:

\exp (0)

1

⇥

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

▶Inverse Laplace transform:

\exp (t) - 1