inverse oflaplace ((2s+1))/((s+1)^2)

⚠Ⓘ 2s is read as the product 2*s: for a power, write 2^s

⇥1) You can decompose into partial fractions:

⇥

\frac{2 s + 1}{{(s + 1)}^{2}} = \frac{A}{s + 1} + \frac{B}{{(s + 1)}^{2}}

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

{(s + 1)}^{2}

, which yields:

⇥

2 s + 1 = A \cdot (s + 1) + B

⇥

2 s + 1 = (A s + A) + B

⇥

2 s + 1 = A s + A + B

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

⇥ [1]:

2 = A

⇥ [2]:

1 = A + B

⇥Solution found [1]:

A = 2

⇥3) Let's solve for B:

1 = A + B

[2] using

A = 2

⇥

2 + B = 1

⇥

B = 1 - 2

⇥Solution found [2]:

B = -1

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

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{2}{s + 1}

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

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

- \frac{1}{{(s + 1)}^{2}}

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

⇥

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

⇥ℹ

-1 \cdot t

- t

‖

-1 \cdot t \exp (- t)

- t \exp (- t)

⇥

\frac{2 \exp (- t)}{1} - \frac{t \exp (- t)}{1}

⇥ℹ

2 \exp (- t)

\frac{2}{\exp (t)}

‖

- t \exp (- t)

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

⇥

\frac{2}{\exp (t)} - \frac{t}{\exp (t)}

▶Inverse Laplace transform:

\frac{2 - t}{\exp (t)}