Form the differential equation having $y=(\sin^{-1}x)^2+A\cos^{-1}x+B$, where $A$ and $B$ are arbitrary constants, as its general solution.

Form the differential equation having y=(sin⁻¹x)²+Acos⁻¹x+B, where A and B are arbitrary constants, as its gen

NCERT Exemplar · Q13

y=(\sin^{-1}x)^2+A\cos^{-1}x+B

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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 two arbitrary constants $A$ and $B$ by successive differentiation. Differentiating twice gives $y' = \frac{2\sin^{-1}x}{\sqrt{1-x^2}} - \frac{A}{\sqrt{1-x^2}}$ and $y'' = \frac{2}{1-x^2} + \frac{x}{(1-x^2)^{3/2}}(A - 2\sin^{-1}x)$ . Substituting $A$ from the first derivative into the second yields the differential equation: $(1-x^2)y'' - xy' = 2$ .

The problem asks us to form the differential equation whose general solution is given. That means we must eliminate the arbitrary constants $A$ and $B$ from the given relation.

Why differentiate?

A general solution with $n$ arbitrary constants corresponds to a differential equation of order $n$ . Here we have two constants ( $A$ and $B$ ), so we need a second-order differential equation. Differentiating the given equation introduces the constants in the derivatives; we then use algebraic elimination to remove them.

Step-by-step solution

1. Write the given equation.

We have:

y = (\sin^{-1}x)^2 + A\cos^{-1}x + B.

2. Differentiate once with respect to $x$ .

Recall:

$\frac{d}{dx}(\sin^{-1}x)^2 = 2\sin^{-1}x \cdot \frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}(B) = 0$

So:

y' = \frac{2\sin^{-1}x}{\sqrt{1-x^2}} - \frac{A}{\sqrt{1-x^2}}.

Factor $\frac{1}{\sqrt{1-x^2}}$ :

y' = \frac{2\sin^{-1}x - A}{\sqrt{1-x^2}}. \qquad(1)

3. Differentiate again to get $y''$ .

Differentiate (1) using the quotient rule (or product rule). Write:

y' = (2\sin^{-1}x - A) \cdot (1-x^2)^{-1/2}.

Let $u = 2\sin^{-1}x - A$ and $v = (1-x^2)^{-1/2}$ . Then:

$u' = \frac{2}{\sqrt{1-x^2}}$
$v' = -\frac{1}{2}(1-x^2)^{-3/2} \cdot (-2x) = \frac{x}{(1-x^2)^{3/2}}$

By the product rule:

y'' = u'v + uv' = \frac{2}{\sqrt{1-x^2}} \cdot \frac{1}{\sqrt{1-x^2}} + (2\sin^{-1}x - A) \cdot \frac{x}{(1-x^2)^{3/2}}.

Simplify the first term:

y'' = \frac{2}{1-x^2} + \frac{x(2\sin^{-1}x - A)}{(1-x^2)^{3/2}}. \qquad(2)

4. Eliminate $A$ using equation (1).

From (1):

$2\sin^{-1}x - A = y'\sqrt{1-x^2}.$ …

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Q.Form the differential equation having $y=(\sin^{-1}x)^2+A\cos^{-1}x+B$ , where $A$ and $B$ are arbitrary constants, as its general solution.

Concept understanding — Eliminating Arbitrary Constant

Eliminating Arbitrary Constants – The Core Idea

Why Differentiate?

The Precise Statement

A Second Example (Two Constants)

Why This Matters …

Step-by-step solution

Method: Forming a second-order DE by eliminating two constants

Steps

Common Mistakes

Q.Form the differential equation having y=(sin⁡−1x)2+Acos⁡−1x+By=(\sin^{-1}x)^2+A\cos^{-1}x+By=(sin−1x)2+Acos−1x+B, where AAA and BBB are arbitrary constants, as its general solution.

Step-by-step solution

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Q.Form the differential equation having $y=(\sin^{-1}x)^2+A\cos^{-1}x+B$ , where $A$ and $B$ are arbitrary constants, as its general solution.

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