Form the differential equation by eliminating A and B in Ax²+By²=1.

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Concept understanding — Eliminating Arbitrary Constant

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:

x + y \frac{dy}{dx} = 0

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

Note

We needed one differentiation because there was one arbitrary constant. In general, $n$ arbitrary constants require $n$ differentiations.

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:

$y = A x + B x^2$
$y' = A + 2B x$
$y'' = 2B$

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

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

y = (y' - x y'') x + \left(\frac{y''}{2}\right) x^2 = x y' - \frac{x^2 y''}{2}

Multiply through by 2 and rearrange:

x^2 y'' - 2x y' + 2y = 0

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

Watch out

A common mistake: trying to eliminate constants by substitution before differentiating enough times. You must differentiate first, then eliminate — substituting too early loses information.

Why This Matters …

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Eliminating two arbitrary constants $A$ and $B$ from $Ax^2+By^2=1$ requires two derivatives (since two constants need two equations to eliminate them). Differentiating twice and solving yields the differential equation $xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0$ .

The core idea: when a relation contains arbitrary constants, each differentiation introduces a new equation linking the constants to derivatives. To eliminate $n$ constants, you need $n$ differentiations (giving $n+1$ equations total, including the original). Here we have two constants $A$ and $B$ , so we differentiate twice.

Why this works: The original equation is a family of curves — each choice of $A$ and $B$ gives a specific curve. The differential equation we seek is the common property shared by all curves in that family, independent of $A$ and $B$ . Differentiating strips away the constants layer by layer, leaving only relationships between $x$ , $y$ , and derivatives.

Let’s go step by step.

1. Start with the given equation

Ax^2 + By^2 = 1

This is our base. $A$ and $B$ are the constants to eliminate.

2. Differentiate once with respect to $x$

Treat $y$ as a function of $x$ . Differentiating term by term:

Derivative of $Ax^2$ is $2Ax$
Derivative of $By^2$ is $B \cdot 2y \cdot \frac{dy}{dx}$ (chain rule)
Derivative of $1$ is $0$

So we get:

2Ax + 2By \frac{dy}{dx} = 0

Divide through by 2:

Ax + By \frac{dy}{dx} = 0 \quad \text{(Equation 1)}

3. Differentiate again (second derivative)

Differentiate Equation 1 with respect to $x$ . Use the product rule on $Ax$ (gives $A$ ) and on $By \frac{dy}{dx}$ :

Derivative of $By \frac{dy}{dx}$ : treat $B$ as constant. Use product rule: $B \left[ \frac{dy}{dx} \cdot \frac{dy}{dx} + y \cdot \frac{d^2y}{dx^2} \right] = B\left[ \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2} \right]$

So differentiating Equation 1 gives:

A + B\left[ \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2} \right] = 0 \quad \text{(Equation 2)}

4. Now we have three equations: original, (1), and (2) — but only two constants to eliminate

We need to eliminate $A$ and $B$ from these. A clean method: solve for $A$ and $B$ from two equations and substitute into the third.

From Equation 1: $Ax = -By \frac{dy}{dx}$ , so $A = -\frac{By}{x} \frac{dy}{dx}$ (provided $x \neq 0$ ).

Substitute this $A$ into Equation 2:

-\frac{By}{x} \frac{dy}{dx} + B\left[ \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2} \right] = 0

Factor $B$ (assuming $B \neq 0$ , otherwise the original equation reduces to $Ax^2=1$ which is a different family — but we want the general case): …

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Q.Form the differential equation by eliminating $A$ and $B$ in $Ax^2+By^2=1$ .

Concept understanding — Eliminating Arbitrary Constant

Eliminating Arbitrary Constants – The Core Idea

Why Differentiate?

The Precise Statement

A Second Example (Two Constants)

Why This Matters …

Method: Eliminating two constants from a conic family

Steps

Common Mistakes

Q.Form the differential equation by eliminating AAA and BBB in Ax2+By2=1Ax^2+By^2=1Ax2+By2=1.

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Q.Form the differential equation by eliminating $A$ and $B$ in $Ax^2+By^2=1$ .

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