Find the inverse Laplace transform of F(s) = 1 / (s2(s+1)(s2+9)) using partial fraction decomposition.

To find the inverse Laplace transform of the given function, we first perform partial fraction decomposition.

The function is:

$$F(s) = \frac{1}{s^2(s+1)(s^2+9)}$$

Step 1: Partial Fraction Decomposition
We set up the partial fraction decomposition as follows:

$$\frac{1}{s^2(s+1)(s^2+9)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1} + \frac{Ds+E}{s^2+9}$$

Multiply both sides by $s^2(s+1)(s^2+9)$:

$$1 = A s(s+1)(s^2+9) + B (s+1)(s^2+9) + C s^2(s^2+9) + (Ds+E) s^2(s+1)$$

To find the coefficients, we can substitute specific values for $s$:

• Set $s=0$: $$1 = A(0) + B(0+1)(0^2+9) + C(0) + (D(0)+E)(0)$$ $$1 = B(1)(9) \implies 9B = 1 \implies B = \frac{1}{9}$$
• Set $s=-1$: $$1 = A(-1)(-1+1)((-1)^2+9) + B(-1+1)((-1)^2+9) + C(-1)^2((-1)^2+9) + (D(-1)+E)(-1)^2(-1+1)$$ $$1 = A(0) + B(0) + C(1)(1+9) + (D(-1)+E)(0)$$ $$1 = 10C \implies C = \frac{1}{10}$$

Now, expand the equation and equate coefficients of powers of $s$:

$$1 = A(s^4+s^3+9s^2+9s) + B(s^3+s^2+9s+9) + C(s^4+9s^2) + (Ds+E)(s^3+s^2)$$ $$1 = A(s^4+s^3+9s^2+9s) + B(s^3+s^2+9s+9) + C(s^4+9s^2) + (Ds^4+Ds^3+Es^3+Es^2)$$ $$1 = (A+C+D)s^4 + (A+B+D+E)s^3 + (9A+B+9C+E)s^2 + (9A+9B)s + 9B$$

Equating coefficients:

• Constant term: $1 = 9B$. This confirms $B = \frac{1}{9}$.
• Coefficient of $s$: $0 = 9A+9B$ Substitute $B = \frac{1}{9}$: $$0 = 9A + 9\left(\frac{1}{9}\right) \implies 0 = 9A + 1 \implies A = -\frac{1}{9}$$
• Coefficient of $s^4$: $0 = A+C+D$ Substitute $A = -\frac{1}{9}$ and $C = \frac{1}{10}$: $$0 = -\frac{1}{9} + \frac{1}{10} + D \implies D = \frac{1}{9} - \frac{1}{10} = \frac{10-9}{90} = \frac{1}{90}$$