The differential equation of the family of curves $x^2+y^2-2ay=0$, where $a$ is arbitrary constant, is: (A) $(x^2-y^2)\frac{dy}{dx}=2xy$ (B) $2(x^2+y^2)\frac{dy}{dx}=xy$ (C) $2(x^2-y^2)\frac{dy}{dx}=xy$ (D) $(x^2+y^2)\frac{dy}{dx}=2xy$

> [!TLDR] > We eliminate the arbitrary constant $a$ by differentiating the given curve equation and then substituting back to remove $a$, obtaining the differential equation $(x^2-y^2)\frac{dy}{dx}=2xy$, which corresponds to option **(A)**. The core idea: a family of curves containing one arbitrary constant (here $a$) corresponds to a first-order differential equation. To find it, we differentiate the given equation once (introducing $\frac{dy}{dx}$) and then use the original equation to eliminate $a$. This yields a relation between $x$, $y$, and $\frac{dy}{dx}$ that holds for every curve in the family — that is the required differential equation. Let’s work through it. 1. **Start with the given family** $$x^2 + y^2 - 2ay = 0$$ Here $a$ is the arbitrary constant. Our goal: remove $a$ by combining this equation with its derivative. 2. **Differentiate both sides with respect to $x$** Remember $y$ is a function of $x$, so we use implicit differentiation: $$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - 2a \frac{d}{dx}(y) = 0$$ $$2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0$$ Divide through by 2: $$x + y\frac{dy}{dx} - a\frac{dy}{dx} = 0$$ 3. **Solve this derivative equation for $a$** Rearranging: $$x + y\frac{dy}{dx} = a\frac{dy}{dx}$$ So $$a = \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}}$$ provided $\frac{dy}{dx} \neq 0$ (which is fine — we’re not at a horizontal tangent point for the general family). 4. **Substitute this $a$ back into the original equation** Original: $x^2 + y^2 - 2ay = 0$ becomes $$x^2 + y^2 - 2\left(\frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}}\right) y = 0$$ 5. **Simplify algebraically** Multiply through by $\frac{dy}{dx}$ to clear the denominator: $$(x^2 + y^2)\frac{dy}{dx} - 2y\left(x + y\frac{dy}{dx}\right) = 0$$ Expand the second term: $$(x^2 + y^2)\frac{dy}{dx} - 2xy - 2y^2\frac{dy}{dx} = 0$$ …

The differential equation of the family of curves x²+y²-2ay=0, where a is arbitrary constant, is: (A) (x²-y²)(

NCERT Exemplar · Q81

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

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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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We eliminate the arbitrary constant $a$ by differentiating the given curve equation and then substituting back to remove $a$ , obtaining the differential equation $(x^2-y^2)\frac{dy}{dx}=2xy$ , which corresponds to option (A).

The core idea: a family of curves containing one arbitrary constant (here $a$ ) corresponds to a first-order differential equation. To find it, we differentiate the given equation once (introducing $\frac{dy}{dx}$ ) and then use the original equation to eliminate $a$ . This yields a relation between $x$ , $y$ , and $\frac{dy}{dx}$ that holds for every curve in the family — that is the required differential equation.

Let’s work through it.

Start with the given family

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

Here $a$ is the arbitrary constant. Our goal: remove $a$ by combining this equation with its derivative.

Differentiate both sides with respect to $x$ Remember $y$ is a function of $x$ , so we use implicit differentiation:

\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - 2a \frac{d}{dx}(y) = 0

2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0

Divide through by 2:

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

Solve this derivative equation for $a$ Rearranging:

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

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

provided $\frac{dy}{dx} \neq 0$ (which is fine — we’re not at a horizontal tangent point for the general family).

Substitute this $a$ back into the original equation Original: $x^2 + y^2 - 2ay = 0$ becomes

x^2 + y^2 - 2\left(\frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}}\right) y = 0

Simplify algebraically Multiply through by $\frac{dy}{dx}$ to clear the denominator:

(x^2 + y^2)\frac{dy}{dx} - 2y\left(x + y\frac{dy}{dx}\right) = 0

Expand the second term:

$(x^2 + y^2)\frac{dy}{dx} - 2xy - 2y^2\frac{dy}{dx} = 0$ …

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Q.The differential equation of the family of curves $x^2+y^2-2ay=0$ , where $a$ is arbitrary constant, is:
(A) $(x^2-y^2)\frac{dy}{dx}=2xy$
(B) $2(x^2+y^2)\frac{dy}{dx}=xy$
(C) $2(x^2-y^2)\frac{dy}{dx}=xy$
(D) $(x^2+y^2)\frac{dy}{dx}=2xy$

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: Form the differential equation by eliminating one arbitrary constant

Steps

Common Mistakes

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