https://file.aiquickdraw.com/mathbot/file/1725433980147axyregxb.jpg

The problem in the image asks you to find the inverse Laplace transform of the given function:

\[
F(s) = \frac{s^2 - 4}{s^3 + 6s^2 + 9s}
\]

To solve this, we need to simplify the given expression and then find the inverse Laplace transform.

### Step 1: Simplify the Function
Let's factor the denominator first:

\[
s^3 + 6s^2 + 9s = s(s^2 + 6s + 9)
\]

The quadratic term can be factored as:

\[
s^2 + 6s + 9 = (s + 3)(s + 3) = (s + 3)^2
\]

Thus, the denominator can be rewritten as:

\[
s(s + 3)^2
\]

So the function becomes:

\[
F(s) = \frac{s^2 - 4}{s(s + 3)^2}
\]

### Step 2: Partial Fraction Decomposition
We can decompose this into partial fractions:

\[
\frac{s^2 - 4}{s(s + 3)^2} = \frac{A}{s} + \frac{B}{s + 3} + \frac{C}{(s + 3)^2}
\]

Multiplying both sides by the common denominator \(s(s + 3)^2\) gives:

\[
s^2 - 4 = A(s + 3)^2 + Bs(s + 3) + Cs
\]

Expanding and solving for coefficients \(A\), \(B\), and \(C\) by equating coefficients of corresponding powers of \(s\) will allow us to find the inverse Laplace transform.

### Step 3: Equate Coefficients
Let's first expand the terms:

\[
A(s + 3)^2 = A(s^2 + 6s + 9)
\]
\[
Bs(s + 3) = Bs^2 + 3Bs
\]
\[
Cs = Cs
\]

Now combining all the terms:

\[
s^2 - 4 = (A + B)s^2 + (6A + 3B + C)s + 9A
\]

Equating the coefficients of \(s^2\), \(s\), and the constant term:

1. Coefficient of \(s^2\): \(A + B = 1\)
2. Coefficient of \(s\): \(6A + 3B + C = 0\)
3. Constant term: \(9A = -4\)

### Step 4: Solve the System of Equations
From the constant term equation:
\[
A = -\frac{4}{9}
\]

Substitute \(A\) into the equation \(A + B = 1\):

\[
-\frac{4}{9} + B = 1 \implies B = 1 + \frac{4}{9} = \frac{13}{9}
\]

Substitute \(A\) and \(B\) into the equation \(6A + 3B + C = 0\):

\[
6\left(-\frac{4}{9}\right) + 3\left(\frac{13}{9}\right) + C = 0
\]

Simplifying:

\[
-\frac{24}{9} + \frac{39}{9} + C = 0 \implies \frac{15}{9} + C = 0 \implies C = -\frac{15}{9} = -\frac{5}{3}
\]

### Step 5: Express the Partial Fractions
The partial fractions become:

\[
F(s) = \frac{-4/9}{s} + \frac{13/9}{s + 3} - \frac{5/3}{(s + 3)^2}
\]

### Step 6: Find the Inverse Laplace Transform
Now, we can find the inverse Laplace transform of each term:

1. \(\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1\)
2. \(\mathcal{L}^{-1}\left\{\frac{1}{s + a}\right\} = e^{-at}\)
3. \(\mathcal{L}^{-1}\left\{\frac{1}{(s + a)^2}\right\} = te^{-at}\)

Applying these:

\[
f(t) = -\frac{4}{9} \cdot 1 + \frac{13}{9}e^{-3t} - \frac{5}{3}te^{-3t}
\]

Thus, the inverse Laplace transform is:

\[
f(t) = -\frac{4}{9} + \frac{13}{9}e^{-3t} - \frac{5}{3}te^{-3t}
\]

Would you like further details on any step? Here are some related questions:

1. How do you factor a cubic polynomial like \(s^3 + 6s^2 + 9s\)?
2. What are partial fraction decompositions, and why are they useful?
3. What is the Laplace transform, and how is it used in differential equations?
4. How can you identify the terms in partial fractions for different Laplace transforms?
5. How do you apply the inverse Laplace transform to solve differential equations?

**Tip:** Always check if a polynomial can be factored or simplified before applying partial fraction decomposition—it simplifies the process significantly.

Find the inverse Laplace transform of the following: F(s) = (s^2 - 4) / (s^3 + 6s^2 + 9s)

Learn how to find the inverse Laplace transform of the function F(s) = (s^2 - 4) / (s^3 + 6s^2 + 9s) using partial fraction decomposition and applying key inverse Laplace transform formulas. This detailed solution is ideal for undergraduate students studying differential equations and transforms.

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