Find the differential equation for the family of curves given by $y = a e^{3x} + b e^{-2x}$ by eliminating the arbitrary constants $a$ and $b$.

(N/A) Given equation: $y = a e^{3x} + b e^{-2x}$ .............$(1)$
Differentiating with respect to $x$:
$y' = 3a e^{3x} - 2b e^{-2x}$ .............$(2)$
Differentiating again with respect to $x$:
$y'' = 9a e^{3x} + 4b e^{-2x}$ .............$(3)$
To eliminate $a$ and $b$,we can use the characteristic equation method or solve the system of equations. From $(1)$ and $(2)$:
$y' + 2y = (3a e^{3x} - 2b e^{-2x}) + 2(a e^{3x} + b e^{-2x}) = 5a e^{3x} \Rightarrow a e^{3x} = \frac{y' + 2y}{5}$
$y' - 3y = (3a e^{3x} - 2b e^{-2x}) - 3(a e^{3x} + b e^{-2x}) = -5b e^{-2x} \Rightarrow b e^{-2x} = \frac{3y - y'}{5}$
Substituting these into $(3)$:
$y'' = 9\left(\frac{y' + 2y}{5}\right) + 4\left(\frac{3y - y'}{5}\right)$
$y'' = \frac{9y' + 18y + 12y - 4y'}{5}$
$y'' = \frac{5y' + 30y}{5}$
$y'' = y' + 6y$
$y'' - y' - 6y = 0$