Q.The differential equation of the family of curves $y=\mathrm{A}e^x+\mathrm{B}e^{-x}$, where $\mathrm{A}$ and $\mathrm{B}$ are arbitrary constants is

A differential equation can be manufactured from any family of curves (or functions) that carries arbitrary constants, by eliminating those constants — and, conversely, verifying that a given expression is a solution of a stated differential equation is the reverse check of the same idea.

The elimination method. Suppose a family of curves is written with $n$ arbitrary constants. To form the differential equation that this whole family satisfies (and that no longer contains any of those constants):

1. Differentiate the defining equation successively $n$ times, producing $n$ new equations.
2. Together with the original equation, that gives $(n+1)$ equations.
3. Eliminate the $n$ arbitrary constants from these $(n+1)$ equations algebraically.
4. The result is a differential equation of order $n$ — order exactly matches the number of constants eliminated: one constant gives a first-order equation, two constants give a second-order equation, and so on.

This is genuinely an elimination problem, not a differentiation recipe alone — after differentiating, the constants are isolated and substituted back (or the several equations are combined) until every trace of $\mathrm A$, $\mathrm B$, $a$, $b$, … is gone and only $x,y$ and derivatives of $y$ remain. …