Q.The differential equation of the family of curves $y^2=4a(x+a)$ is:
(A) $y^2=4\frac{dy}{dx}\left(x+\frac{dy}{dx}\right)$
(B) $2y\frac{dy}{dx}=4a$
(C) $y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=0$
(D) $2x\frac{dy}{dx}+y\left(\frac{dy}{dx}\right)^2-y=0$

Eliminating Arbitrary Constants – The Core Idea

Consider a family of curves — say all circles centred at the origin: $x^2 + y^2 = r^2$. Here $r$ is an arbitrary constant: each value gives one specific circle, but the whole family shares the same shape.

Suppose you're asked: "What differential equation does every member satisfy?" You need to eliminate the arbitrary constant $r$ to get an equation that holds for all such circles, regardless of $r$.

That is the goal: start with an equation containing arbitrary constants, differentiate enough times to remove them, and obtain a differential equation representing the whole family.

Why Differentiate?

Arbitrary constants are constants — their derivative is zero. So differentiating either makes the constant disappear or lets you eliminate it by combining the original equation with its derivatives.

Differentiate $x^2 + y^2 = r^2$ with respect to $x$: $\;2x + 2y \frac{dy}{dx} = 0$. Divide by 2:

The constant $r$ is gone. The differential equation $x + y y' = 0$ is satisfied by every circle centred at the origin.

The Precise Statement

Eliminating arbitrary constants means: given an equation involving $x$, $y$, and $n$ arbitrary constants, differentiate it $n$ times (with respect to the independent variable) and then algebraically eliminate the $n$ constants from the system of $n+1$ equations (the original plus the $n$ derivatives). The result is an $n$-th order ordinary differential equation representing the entire family of curves.

A Second Example (Two Constants)

Take all curves $y = A x + B x^2$, where $A$ and $B$ are arbitrary constants.

Differentiate once: $\;y' = A + 2B x$

Differentiate again: $\;y'' = 2B$

Now we have three equations:

1. $y = A x + B x^2$
2. $y' = A + 2B x$
3. $y'' = 2B$

From (3), $B = \frac{y''}{2}$. Substitute into (2): $y' = A + x y''$, so $A = y' - x y''$.

Substitute $A$ and $B$ into (1):

Multiply through by 2 and rearrange:

This second-order ODE is satisfied by every curve $y = A x + B x^2$, no matter what $A$ and $B$ are.

Why This Matters …