The differential equation for $y=A\cos\alpha x+B\sin\alpha x$, where $A$ and $B$ are arbitrary constants, is: (A) $\frac{d^2y}{dx^2}-\alpha^2 y=0$ (B) $\frac{d^2y}{dx^2}+\alpha^2 y=0$ (C) $\frac{d^2y}{dx^2}+\alpha y=0$ (D) $\frac{d^2y}{dx^2}-\alpha y=0$

The differential equation for y=Acosα x+Bsinα x, where A and B are arbitrary constants, is: (A) (d²y)/(dx²)-α²

NCERT Exemplar · Q60

y=A\cos\alpha x+B\sin\alpha x

Madhya Pradesh MpbseMCQ· 1mImportance★★★★★

Appeared in past exams:MHT-CET 2021· Set pcm-2021-09-23-E· 2mreworded

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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 function is a linear combination of $\cos\alpha x$ and $\sin\alpha x$ , which are solutions of the second‑order ODE $\frac{d^2y}{dx^2}+\alpha^2 y=0$ . Differentiating twice and substituting shows that option (B) is correct.

We start with

y = A\cos\alpha x + B\sin\alpha x,

where $A$ and $B$ are arbitrary constants. The task is to eliminate these constants and obtain a differential equation that $y$ satisfies, no matter what $A$ and $B$ are.

The key idea: a relation involving two arbitrary constants will generally require two differentiations to remove them. Each differentiation introduces a new equation, and we combine them to eliminate $A$ and $B$ .

First derivative Differentiate $y$ with respect to $x$ :

\frac{dy}{dx} = -A\alpha\sin\alpha x + B\alpha\cos\alpha x.

Second derivative Differentiate again:

\frac{d^2y}{dx^2} = -A\alpha^2\cos\alpha x - B\alpha^2\sin\alpha x.

Factor $-\alpha^2$ out of the right‑hand side:

\frac{d^2y}{dx^2} = -\alpha^2\bigl(A\cos\alpha x + B\sin\alpha x\bigr).

Recognise the original function The bracket is exactly $y$ :

\frac{d^2y}{dx^2} = -\alpha^2 y.

Rearrange into standard form Bring all terms to one side: $\frac{d^2y}{dx^2} + \alpha^2 y = 0.$ …

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Q.The differential equation for $y=A\cos\alpha x+B\sin\alpha x$ , where $A$ and $B$ are arbitrary constants, is:
(A) $\frac{d^2y}{dx^2}-\alpha^2 y=0$
(B) $\frac{d^2y}{dx^2}+\alpha^2 y=0$
(C) $\frac{d^2y}{dx^2}+\alpha y=0$
(D) $\frac{d^2y}{dx^2}-\alpha y=0$

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 the DE of $y=A\cos\alpha x+B\sin\alpha x$

Steps

Common Mistakes

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