CBSE Board Examination 2026 · Section D · 5 marks

Q34: If $x = \cos t$ ⁠, $y = \cos mt$ ⁠, prove $(1 - x^2)y'' - xy' + m^2y = 0$

Board 2026Ch 5 · Continuity and Differentiability

The question

If $x = \cos t$ ⁠, $y = \cos mt$ ⁠, prove that

(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + m^2y = 0.

Exam answer

\frac{dx}{dt} = -\sin t,\quad \frac{dy}{dt} = -m\sin mt

Why?

$x$ and $y$ are both given in terms of a parameter $t$ ⁠, so first differentiate each with respect to $t$ ⁠.

By the chain rule, the inner function $mt$ has derivative $m$ ⁠:

\frac{d}{dt}\cos mt = -m\sin mt

\therefore\ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{m\sin mt}{\sin t}

Why?

For parametric equations,

\frac{dy}{dx} = \frac{dy}{dt}\div\frac{dx}{dt}

The two minus signs cancel.

\begin{aligned} \frac{dy}{dx} &= \frac{-m\sin mt}{-\sin t} \\ &= \frac{m\sin mt}{\sin t} \end{aligned}

Then multiply both sides by $\sin t$ to get the next line.

\Rightarrow \sin t\,\frac{dy}{dx} = m\sin mt

Why?

Clearing the fraction first is a good idea: the next derivative is then a product rule instead of a quotient rule, and the working is shorter.

Differentiating both sides w.r.t. $x$ ⁠,

\begin{aligned} &\sin t\,\frac{d^2y}{dx^2} + \cos t\,\frac{dt}{dx}\cdot\frac{dy}{dx} \\ &= m^2\cos mt\,\frac{dt}{dx} \end{aligned}

Why?

The left side is a product, $\sin t$ times $\frac{dy}{dx}$ ⁠, so use the product rule.

$\sin t$ and $m\sin mt$ are functions of $t$ ⁠, and $t$ is a function of $x$ ⁠. So by the chain rule

\begin{aligned} \frac{d}{dx}\sin t &= \cos t\,\frac{dt}{dx} \\ \frac{d}{dx}(m\sin mt) &= m^2\cos mt\,\frac{dt}{dx} \end{aligned}

For the left side, take $u = \sin t$ and $v = \frac{dy}{dx}$ ⁠:

\begin{aligned} u' &= \cos t\,\frac{dt}{dx} \\ v' &= \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2} \\[4pt] (uv)' &= uv' + vu' \\ &= \sin t\,\frac{d^2y}{dx^2} + \frac{dy}{dx}\cos t\,\frac{dt}{dx} \end{aligned}

For the right side, $m\sin mt$ ⁠: the derivative of $\sin mt$ with respect to $t$ is $m\cos mt$ ⁠, so $m \cdot m\cos mt = m^2\cos mt$ ⁠, times $\frac{dt}{dx}$ ⁠.

Putting $\frac{dt}{dx} = -\frac{1}{\sin t}$ and multiplying both sides by $\sin t$ ⁠,

\begin{aligned} &\sin t\,\frac{d^2y}{dx^2} - \frac{\cos t}{\sin t}\cdot\frac{dy}{dx} \\ &= -\frac{m^2\cos mt}{\sin t} \end{aligned}

\sin^2t\,\frac{d^2y}{dx^2} - \cos t\,\frac{dy}{dx} = -m^2\cos mt

Why?

$\frac{dt}{dx}$ is the reciprocal of $\frac{dx}{dt} = -\sin t$ ⁠. Putting it in gives the first line.

\begin{aligned} \frac{dt}{dx} &= \frac{1}{dx/dt} = \frac{1}{-\sin t} = -\frac{1}{\sin t} \\[4pt] &\cos t\cdot\left(-\frac{1}{\sin t}\right)\cdot\frac{dy}{dx} \\ &= -\frac{\cos t}{\sin t}\cdot\frac{dy}{dx} \\[4pt] &m^2\cos mt\cdot\left(-\frac{1}{\sin t}\right) \\ &= -\frac{m^2\cos mt}{\sin t} \end{aligned}

Multiplying every term by $\sin t$ clears the denominators. The first term becomes $\sin^2t\,\frac{d^2y}{dx^2}$ ⁠, and in the other two terms $\sin t$ cancels.

\begin{aligned} \sin t \cdot \sin t\,\frac{d^2y}{dx^2} &= \sin^2t\,\frac{d^2y}{dx^2} \\ \sin t \cdot \frac{\cos t}{\sin t}\cdot\frac{dy}{dx} &= \cos t\,\frac{dy}{dx} \\ \sin t \cdot \frac{m^2\cos mt}{\sin t} &= m^2\cos mt \end{aligned}

As $\sin^2t = 1 - \cos^2t = 1 - x^2$ ⁠, $\cos t = x$ and $\cos mt = y$ ⁠,

Why?

The result must be in $x$ and $y$ ⁠, so replace every expression in $t$ ⁠. Use

\sin^2t + \cos^2t = 1

together with the given $x = \cos t$ and $y = \cos mt$ ⁠.

\begin{aligned} \sin^2t &= 1 - \cos^2t = 1 - x^2 \\ \cos t &= x \\ -m^2\cos mt &= -m^2y \end{aligned}

(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = -m^2y

Why?

Add $m^2y$ to both sides to get the required form with 0 on the right.

$\therefore\ (1 - x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} + m^2y = 0$ ⁠. Hence proved.

Where this comes from

Chapter 5, Continuity and Differentiability: Second derivative of a parametric function.

Question type: Parametric form + Second derivative (combines 2 types)

How to spot it: click to see

Type 8. Parametric form: $x$ and $y$ given through $t$ or $\theta$

Spot it: two equations, one for $x$ and one for $y$ ⁠, both in the same third letter.

Here: First, $x = \cos t$ and $y = \cos mt$ share $t$ ⁠: divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to get $\frac{m\sin mt}{\sin t}$ ⁠, and clear the fraction: $\sin t\,\frac{dy}{dx} = m\sin mt$ ⁠.

Type 9. Second derivative: differentiate twice, then prove

Spot it: $\frac{d^2y}{dx^2}$ or $y''$ in the question, often "prove that $\frac{d^2y}{dx^2} + y = 0$ ".

Here: Then differentiate that again with respect to $x$ ⁠, using $\frac{dt}{dx} = -\frac{1}{\sin t}$ ⁠, and swap $\sin^2 t$ ⁠, $\cos t$ and $\cos mt$ for $1 - x^2$ ⁠, $x$ and $y$ ⁠.

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Continuity and Differentiability in past papers: question 2 of 35 · see them all

Q34: If x=cos⁡tx = \cos tx=cost⁠, y=cos⁡mty = \cos mty=cosmt⁠, prove (1−x2)y′′−xy′+m2y=0(1 - x^2)y'' - xy' + m^2y = 0(1−x2)y′′−xy′+m2y=0

Q34: If $x = \cos t$ ⁠, $y = \cos mt$ ⁠, prove $(1 - x^2)y'' - xy' + m^2y = 0$