I was solving a problem but I am stuck at it. Here is the question :
$\frac{7s^2+9s+3}{(s^2-12s+40)(s^2+9)}$ Find inverse Laplace transform.
I performed these operation :
$\frac{As+B}{(s^2-12s+40)}$ + $\frac{Cs+D}{(s^2+9)}$ =$\frac{7s^2+9s+3}{(s^2-12s+40)(s^2+9)}$
After calculations, I get 2 different values of B. one of them is $\frac{31}{4}$ other one is $\frac{33}{9}$ . Is there a problem about this question?
Multiplying, we get $$(As+B)(s^2+9) + (Cs+D)(s^2-12s+40) = 7s^2 + 9s + 3.$$ Expanding and collecting terms, we obtain \begin{align} 7s^2 + 9s+3 &= A s^3+9 A s+B s^2+9 B+C s^3-12 C s^2+40 C s+D s^2-12 D s+40 D \\ &= (A+C)s^3 + (B-12C+D)s^2 + (9A + 40C -12D)s + (9B + 40D). \end{align} This yields the linear system of equations $$ \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & -12 & 1 \\ 9 & 0 & 40 & -12 \\ 0 & 9 & 0 & 40 \end{bmatrix} \begin{bmatrix} A \\ B \\ C \\ D \end{bmatrix} = \begin{bmatrix} 0 \\ 7 \\ 9 \\ 3 \end{bmatrix}. $$ It is easy to put this into row echelon form (multiply the first and second rows by $9$) and solve numerically from here.