inverse laplace (-5)/(s(s^2-5s+4))

⇥1) Let's factor in

ℝ

s \cdot (s^{2} - 5 s + 4)

⇥Let's factor in

ℝ

(s^{2} - 5 s + 4)

⇥Given the quadratic form

a s^{2} + b s + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -5

and

c = 4

⇥

Δ = {(-5)}^{2} - 4 \cdot 1 \cdot 4

⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-5)}^{2}

(-5) \cdot (-5)

25

⇥

Δ = 25 - 4 \cdot 4

⇥ⓘ Multiplications are done before additions and subtractions:

4 \cdot 4

16

⇥

Δ = 25 - 16

⇥ⓘ Start from

25

and move back

16

on the number line:

25 - 16

9

⇥

Δ = 9

⇥As

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

s_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-5) - \sqrt{9}}{2 \cdot 1}

⇥

s_{1} = \frac{5 - \sqrt{9}}{2}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

⇥

s_{1} = \frac{5 - 3}{2}

⇥ⓘ The numerator and the denominator are computed before the division:

5 - 3

2

⇥

s_{1} = \frac{2}{2}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

⇥

s_{1} = 1

⇥The quadratic formula gives

s_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-5) + \sqrt{9}}{2 \cdot 1}

⇥

s_{2} = \frac{5 + \sqrt{9}}{2}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

⇥

s_{2} = \frac{5 + 3}{2}

⇥ⓘ The numerator and the denominator are computed before the division:

5 + 3

8

⇥

s_{2} = \frac{8}{2}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{8}{2}

4

⇥

s_{2} = 4

⇥Factors identified:

(s - 1) \cdot (s - 4)

⇥2) You can perform a partial fraction decomposition:

⇥

- \frac{5}{s \cdot (s - 1) \cdot (s - 4)} = \frac{A}{s} + \frac{B}{s - 1} + \frac{C}{s - 4}

⇥For the simple factor:

s

and its root:

0

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

s \cdot (s^{2} - 5 s + 4)

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

⇥3) Let's solve in

ℝ

(0 - 1) \cdot (0 - 4) \cdot A = -5

⇥ⓘ Adding 0 changes nothing:

0 - 1

-1

0 - 4

-4

■Two minus signs cancel out, the product of two negative numbers is positive:

-1 \cdot (-4)

4

⇥

4 A = -5

⇥ⓘ Both sides are divided by

4

⇥Solution obtained:

A = - \frac{5}{4}

⇥For the simple factor:

s - 1

and its root:

1

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

s \cdot (s^{2} - 5 s + 4)

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

⇥4) Let's solve in

ℝ

1 \cdot (1 - 4) \cdot B = -5

⇥

(1 - 4) \cdot B = -5

⇥ⓘ Parentheses are computed first:

1 - 4

-3

⇥

- 3 B = -5

⇥

3 B = 5

⇥ⓘ Both sides are divided by

3

⇥Solution obtained:

B = \frac{5}{3}

⇥For the simple factor:

s - 4

and its root:

4

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

s \cdot (s^{2} - 5 s + 4)

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

⇥5) Let's solve in

ℝ

4 \cdot (4 - 1) \cdot C = -5

⇥ⓘ Parentheses are computed first:

4 - 1

3

⇥

4 \cdot 3 C = -5

⇥ⅰ

4 \cdot 3

12

⇥

12 C = -5

⇥ⓘ Both sides are divided by

12

⇥Solution found:

C = - \frac{5}{12}

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

⇥

- \frac{5}{s \cdot (s - 1) \cdot (s - 4)} = - \frac{5}{4 s} + \frac{5}{3 \cdot (s - 1)} - \frac{5}{12 \cdot (s - 4)}

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

Before applying the formula, normalize the pole:

- \frac{5}{4 s} = - \frac{\frac{5}{4}}{s}

The inverse Laplace formula for

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

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

; it gives for

- \frac{\frac{5}{4}}{s}

\frac{- \frac{5}{4} \cdot \exp (0 \cdot t)}{0!}

Before applying the formula, normalize the pole:

\frac{5}{3 \cdot (s - 1)} = \frac{\frac{5}{3}}{s - 1}

The inverse Laplace formula for

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

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

; it gives for

\frac{\frac{5}{3}}{s - 1}

\frac{\frac{5}{3} \cdot \exp (t)}{0!}

Before applying the formula, normalize the pole:

- \frac{5}{12 \cdot (s - 4)} = - \frac{\frac{5}{12}}{s - 4}

The inverse Laplace formula for

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

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

; it gives for

- \frac{\frac{5}{12}}{s - 4}

\frac{- \frac{5}{12} \cdot \exp (4 t)}{0!}

⇥

\frac{- \frac{5}{4} \cdot \exp (0 \cdot t)}{0!} + \frac{\frac{5}{3} \cdot \exp (t)}{0!} + \frac{- \frac{5}{12} \cdot \exp (4 t)}{0!}

⇥ⓘ 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

■Multiplying a fraction by a term multiplies its numerator:

\frac{5}{3} \cdot \exp (t)

\frac{5 \exp (t)}{3}

- \frac{5}{12} \cdot \exp (4 t)

- \frac{5 \exp (4 t)}{12}

■By convention, the factorial of 0 is 1:

0!

1

⇥

- \frac{\frac{5}{4}}{0!} + \frac{\frac{5 \exp (t)}{3}}{1} - \frac{\frac{5 \exp (4 t)}{12}}{1}

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

0!

1

⇥

- \frac{\frac{5}{4}}{1} + \frac{5 \exp (t)}{3} - \frac{5 \exp (4 t)}{12}

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators:

- \frac{5}{4} - \frac{5 \exp (4 t)}{12}

\frac{3 \cdot (-5) - 5 \exp (4 t)}{12}

⇥

\frac{3 \cdot (-5) - 5 \exp (4 t)}{12} + \frac{5 \exp (t)}{3}

⇥ⓘ Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

3 \cdot (-5)

-15

■The minus sign is factored out, changing the sign of each term:

- 15 - 5 \exp (4 t)

- (15 + 5 \exp (4 t))

⇥

- \frac{15 + 5 \exp (4 t)}{12} + \frac{5 \exp (t)}{3}

▶Inverse Laplace transform:

\frac{5 \exp (t)}{3} - \frac{5 \exp (4 t) + 15}{12}