$\tan^{-1}x+\tan^{-1}y=c$ is the general solution of the differential equation: (A) $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$ (B) $\frac{dy}{dx}=\frac{1+x^2}{1+y^2}$ (C) $(1+x^2)\,dy+(1+y^2)\,dx=0$ (D) $(1+x^2)\,dx+(1+y^2)\,dy=0$

tan⁻¹x+tan⁻¹y=c is the general solution of the differential equation: (A) (dy)/(dx)=(1+y²)/(1+x²) (B) (dy)/(dx

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\tan^{-1}x+\tan^{-1}y=c

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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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The given solution $\tan^{-1}x+\tan^{-1}y=c$ implies $\frac{dy}{dx} = -\frac{1+y^2}{1+x^2}$ , which matches option (C) after rearranging.

We have a general solution and need to find which differential equation it satisfies. The key is to differentiate the given relation implicitly and then rearrange to match one of the options.

1. Start with the given solution

The relation is:

\tan^{-1}x + \tan^{-1}y = c

where $c$ is an arbitrary constant. This is a family of curves — the general solution of some first-order differential equation.

2. Differentiate both sides with respect to $x$

Since $y$ is a function of $x$ , we differentiate implicitly:

\frac{d}{dx}\left(\tan^{-1}x\right) + \frac{d}{dx}\left(\tan^{-1}y\right) = \frac{d}{dx}(c)

The derivative of $\tan^{-1}x$ is $\frac{1}{1+x^2}$ . For $\tan^{-1}y$ , we use the chain rule:

\frac{d}{dx}(\tan^{-1}y) = \frac{1}{1+y^2} \cdot \frac{dy}{dx}

And the derivative of a constant $c$ is $0$ . So we get:

\frac{1}{1+x^2} + \frac{1}{1+y^2} \cdot \frac{dy}{dx} = 0

Tip

This step is the heart of the method: when a solution is given implicitly, differentiate term-by-term and treat $y$ as $y(x)$ . The constant vanishes, giving the differential equation directly.

3. Solve for $\frac{dy}{dx}$

Bring the first term to the other side:

\frac{1}{1+y^2} \cdot \frac{dy}{dx} = -\frac{1}{1+x^2}

Multiply both sides by $1+y^2$ :

\frac{dy}{dx} = -\frac{1+y^2}{1+x^2}

This is the differential equation satisfied by the given solution.

4. Match with the options

Option (A) is $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ — missing the negative sign.

Option (B) is $\frac{dy}{dx} = \frac{1+x^2}{1+y^2}$ — reciprocal and wrong sign. …

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Q. $\tan^{-1}x+\tan^{-1}y=c$ is the general solution of the differential equation:
(A) $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$
(B) $\frac{dy}{dx}=\frac{1+x^2}{1+y^2}$
(C) $(1+x^2)\,dy+(1+y^2)\,dx=0$
(D) $(1+x^2)\,dx+(1+y^2)\,dy=0$

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: Forming a DE by Differentiating an Inverse-Trig Relation

Steps

Common Mistakes

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