Q.Form the differential equation representing the family of curves $y = Ae^{2x} + Be^{-3x}$, where $A, B$ are arbitrary constants, and state its order and degree.

Given $y=Ae^{2x}+Be^{-3x}$. Differentiating: $y'=2Ae^{2x}-3Be^{-3x}$, $y''=4Ae^{2x}+9Be^{-3x}$.

Eliminate $A,B$: $y'+3y = 2Ae^{2x}-3Be^{-3x}+3Ae^{2x}+3Be^{-3x}=5Ae^{2x} \implies Ae^{2x}=\dfrac{y'+3y}5$.

Also $y'-2y = 2Ae^{2x}-3Be^{-3x}-2Ae^{2x}-2Be^{-3x}=-5Be^{-3x} \implies Be^{-3x}=\dfrac{2y-y'}5$.

Substitute into $y''$: $y'' = 4Ae^{2x}+9Be^{-3x} = \dfrac{4(y'+3y)}5+\dfrac{9(2y-y')}5 = \dfrac{4y'+12y+18y-9y'}5 = \dfrac{-5y'+30y}5 = -y'+6y$.

So the differential equation is

which has order $2$ (highest derivative $y''$) and degree $1$ (that derivative appears to the first power).

Verification. Substituting $y=Ae^{2x}+Be^{-3x}$ back: $y''+y'-6y = (4Ae^{2x}+9Be^{-3x})+(2Ae^{2x}-3Be^{-3x})-6(Ae^{2x}+Be^{-3x}) = (4+2-6)Ae^{2x}+(9-3-6)Be^{-3x} = 0$, identically, for every $A,B$.

Differentiate the family the same number of times as there are arbitrary constants (twice, for two constants), then eliminate the constants algebraically between $y,y',y''$ to obtain a differential equation free of $A,B$.

Trying to eliminate $A,B$ by inspection instead of systematically solving two linear combinations ($y'+3y$ and $y'-2y$) for $Ae^{2x}$ and $Be^{-3x}$; arithmetic slips combining the fractions.