To find the inverse Laplace transform of $F(s) = \frac{10}{s (s + 2)(s + 3)^2}$, we can use the method of partial fraction decomposition to express $F(s)$ in a form where we can easily apply the inverse Laplace transform.

Step 1: Partial Fraction Decomposition

We start by writing: $\frac{10}{s (s + 2)(s + 3)^2} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+3} + \frac{D}{(s+3)^2}$

To find the coefficients $A, B, C,$ and $D$, we multiply both sides by the denominator $s (s + 2)(s + 3)^2$ to get: $10 = A(s + 2)(s + 3)^2 + Bs(s + 3)^2 + Cs(s + 2)(s + 3) + Ds(s (s + 2))$

Now we will solve for $A, B, C,$ and $D$ by substituting convenient values for $s$.

Solve for $A$:

Let $s = 0$: $10 = A(0 + 2)(0 + 3)^2$ $10 = 18A$ $A = \frac{10}{18} = \frac{5}{9}$

Solve for $B$:

Let $s = -2$: $10 = B(-2)(-2 + 3)^2$ $10 = B(-2)(1)^2$ $10 = -2B$ $B = -5$

Solve for $C$:

Let $s = -3$: $10 = C(-3)(-3 + 2)(-3 + 3) + D(-3)(-3)(-3 + 2)$ $10 = C(-3)(-1)(0) + D(-3)(-3)(-1)$ $10 = 9D$ $D = \frac{10}{9}$

Solve for $D$:

To find $C$, we use any value of $s$ other than those previously used. Let's substitute $s = 1$ into the equation: $10 = A(1 + 2)(1 + 3)^2 + B(1)(1 + 3)^2 + C(1)(1 + 2)(1 + 3) + D(1)(1 + 2)$ $10 = A(3)(4)^2 + B(1)(4)^2 + C(1)(3)(4) + D(1)(3)$ $10 = 48A + 16B + 12C + 3D$ $10 = 48 \left(\frac{5}{9}\right) + 16(-5) + 12C + 3\left(\frac{10}{9}\right)$ $10 = \frac{240}{9} - 80 + 12C + \frac{30}{9}$ $10 = 26.6667 - 80 + 12C + 3.3333$ $10 = -50 + 12C$ $60 = 12C$ $C = 5$

Step 2: Inverse Laplace Transform

Now that we have the coefficients, we can rewrite $F(s)$ as: $F(s) = \frac{5/9}{s} - \frac{5}{s+2} + \frac{5}{s+3} + \frac{10/9}{(s+3)^2}$

We can now find the inverse Laplace transform of each term separately using standard Laplace transform pairs:

1. $\mathcal{L}^{-1} \left\{ \frac{5/9}{s} \right\} = \frac{5}{9} \cdot 1 = \frac{5}{9}$
2. $\mathcal{L}^{-1} \left\{ \frac{5}{s+2} \right\} = 5 e^{-2t}$
3. $\mathcal{L}^{-1} \left\{ \frac{5}{s+3} \right\} = 5 e^{-3t}$
4. $\mathcal{L}^{-1} \left\{ \frac{10/9}{(s+3)^2} \right\} = \frac{10}{9} t e^{-3t}$

Combining these, we get: $\mathcal{L}^{-1} \left\{ F(s) \right\} = \frac{5}{9} - 5 e^{-2t} + 5 e^{-3t} + \frac{10}{9} t e^{-3t}$

Thus, the inverse Laplace transform of $\frac{10}{s (s + 2)(s + 3)^2}$ is: $f(t) = \frac{5}{9} - 5 e^{-2t} + 5 e^{-3t} + \frac{10}{9} t e^{-3t}$

Do you want details or have any questions?

Here are some related questions you might find interesting:

1. What is the inverse Laplace transform of $\frac{1}{s^2 + s + 1}$?
2. How do you find the partial fraction decomposition of a rational function?
3. What is the Laplace transform of $t e^{at}$?
4. How do you compute the inverse Laplace transform of $\frac{1}{s(s+1)(s+2)}$?
5. What is the significance of the Heaviside function in Laplace transforms?
6. How is the convolution theorem used in Laplace transforms?
7. What are the properties of the Laplace transform that make it useful for solving differential equations?
8. Can you provide an example of using Laplace transforms to solve a system of differential equations?

Tip: When performing partial fraction decomposition, always verify your coefficients by substituting back into the original equation to ensure accuracy.