inverse laplace (2s+1)/(s^3) — step-by-step solution

⇥1) You can decompose into partial fractions:

⇥

\frac{2 s + 1}{s^{3}} = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{s^{3}}

⇥To find the remaining unknowns, you multiply both sides by the denominator:

s^{3}

which yields:

⇥

2 s + 1 = A s^{2} + B s + C

⇥2) By identifying the different terms, you obtain the system of equations:

⇥ [1]:

0 = A

⇥ [2]:

2 = B

⇥ [3]:

1 = C

⇥Solution obtained [1]:

A = 0

⇥3) Solving for B:

2 = B

[2]

⇥Solution obtained [2]:

B = 2

⇥4) Let's solve for C:

1 = C

[3]

⇥Solution found [3]:

C = 1

⇥5) Finally, you can substitute the values of the unknowns into the initial partial fraction decomposition:

⇥

\frac{2 s + 1}{s^{3}} = \frac{2}{s^{2}} + \frac{1}{s^{3}}

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{2}{s^{2}}

\frac{2 t \exp (0 \cdot t)}{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{1}{s^{3}}

\frac{t^{2} \exp (0 \cdot t)}{2!}

⇥

\frac{2 t \exp (0 \cdot t)}{1!} + \frac{t^{2} \exp (0 \cdot t)}{2!}

⇥ⓘ 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

■The factorial of a number is the product of all the integers from 1 to that number:

1!

1

2!

2

⇥

\frac{2 t}{1} + \frac{t^{2}}{2}

⇥

2 t + \frac{t^{2}}{2}

▶Inverse Laplace transform:

\frac{t^{2}}{2} + 2 t