Formation of a Differential Equation for Simple Cases

“Forming” a differential equation reverses the idea of a general solution: starting from a family of curves given by an equation containing one or more arbitrary constants, the task is to find a differential equation — free of those constants entirely — that every member of the family satisfies.

The governing rule: a family of curves with exactly $n$ independent arbitrary constants gives rise to a differential equation of order $n$. This is why the method always differentiates exactly as many times as there are constants to eliminate — differentiating fewer times leaves a constant behind; differentiating more times is unnecessary extra work.

Method, one arbitrary constant:

1. Differentiate the given relation once with respect to $x$.
2. Solve the original relation (or the derivative relation, whichever is simpler) for the constant.
3. Substitute this expression for the constant back wherever it still appears, producing a first-order equation with no constant left in it.

Illustration: for the family $y = cx^2$ (one constant $c$), differentiating once gives $\dfrac{dy}{dx} = 2cx$. From the original relation, $c = \dfrac{y}{x^2}$ (for $x \neq 0$); substituting, $\dfrac{dy}{dx} = 2x \cdot \dfrac{y}{x^2} = \dfrac{2y}{x}$, giving the constant-free differential equation $x\dfrac{dy}{dx} = 2y$.

Method, two arbitrary constants:

1. Differentiate the given relation twice, producing three equations in total (the original relation plus its first and second derivatives).
2. Use the two derivative equations to express both constants in terms of $x$, $y$, $\dfrac{dy}{dx}$, and $\dfrac{d^2y}{dx^2}$.
3. Substitute both expressions back into the original relation and simplify — the two constants cancel out completely, leaving a second-order differential equation.

Illustration: for the family $y = Ax + Bx^2$ (two constants $A, B$): differentiating once, $\dfrac{dy}{dx} = A + 2Bx$; differentiating again, $\dfrac{d^2y}{dx^2} = 2B$, so $B = \dfrac{1}{2}\dfrac{d^2y}{dx^2}$. From the first derivative, $A = \dfrac{dy}{dx} - 2Bx = \dfrac{dy}{dx} - x\dfrac{d^2y}{dx^2}$. Substituting both into the original relation, $y = \left(\dfrac{dy}{dx} - x\dfrac{d^2y}{dx^2}\right)x + \dfrac{1}{2}\dfrac{d^2y}{dx^2}x^2$, which simplifies to $x^2\dfrac{d^2y}{dx^2} - 2x\dfrac{dy}{dx} + 2y = 0$ — a second-order equation, matching the two constants that were eliminated. …