Q34

5 marksLong AnswerSection D

If $x = \cos t, y = \cos mt$ , prove that $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + m^2 y = 0$ .

Q: What identity converts $\sin^2t$ into a function of x?

Since $x=\cos t$, $\sin^2t = 1-\cos^2t = 1-x^2$, which is exactly the coefficient appearing in the required equation.

Continuity and Differentiability

Second Order Derivatives of Parametric Functions

Official Answer

This is a proof question — differentiate the parametric functions with respect to the parameter t, form dy/dx in terms of t, then differentiate again to reach the required second-order relation.

Setting up

$x = \cos t \implies dx/dt = -\sin t$ .
$y = \cos mt \implies dy/dt = -m\sin mt$ .
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = m\sin(mt)/\sin t$ , so $\sin t \cdot (dy/dx) = m\sin(mt)$ .

Differentiating again

Differentiating $\sin t\cdot y_1 = m\sin(mt)$ with respect to x (using $dt/dx = -1/\sin t$ ) gives, after simplification, $-x\cdot y_1 + (1-x^2)\cdot y_2 = -m^2 y$ (where $y_1=\frac{dy}{dx}$ , $y_2=\frac{d^2y}{dx^2}$ ).

Conclusion

Rearranging gives exactly $(1-x^2)y_2 - x y_1 + m^2y = 0$ , which is the required result — hence proved.

parametric differentiationsecond order derivativechain ruleChebyshev differential equationimplicit differentiationd2y/dx2

Marking Scheme

11 mark: correctly finding $dx/dt = -\sin t$ and $dy/dt = -m\sin(mt)$ .
21 mark: correctly forming $\frac{dy}{dx} = m\sin(mt)/\sin t$ , i.e. $\sin t \cdot y_1 = m\sin(mt)$ .
32 marks: correctly differentiating implicitly w.r.t. x (using $dt/dx = -1/\sin t$ ) to obtain $-x y_1 + (1-x^2) y_2$ on the LHS.
41 mark: correctly simplifying the RHS to $-m^2y$ (using $\cos(mt) = y$ ) and concluding $(1-x^2)y_2 - xy_1 + m^2y = 0$ .

Hint

Differentiate x and y w.r.t. t separately, form $\frac{dy}{dx} = m\sin(mt)/\sin t$ , then differentiate $\sin t\cdot(dy/dx) = m\sin(mt)$ implicitly w.r.t. x using $dt/dx = -1/\sin t$ .

Quick Oral Answer

Since $x=\cos t$ and $y=\cos mt$ , $\frac{dy}{dx} = m\sin(mt)/\sin t$ ; differentiating this again with respect to x and simplifying using $\sin^2t=1-x^2$ and $\cos(mt)=y$ gives exactly $(1-x^2)y''-xy'+m^2y=0$ .

Analysis & Explanation

This is a classical parametric second-derivative proof that appears repeatedly in CBSE board papers because it elegantly combines chain rule, product rule, and trigonometric identities.

Concept

When both x and y are given in terms of a parameter t, $\frac{dy}{dx}$ is found as $(dy/dt)\div(dx/dt)$ — never differentiate y directly with respect to x.
The key algebraic trick here is recognising that $\sin^2t = 1-\cos^2t = 1-x^2$ , so $\sqrt{1-x^2}$ can replace sin t once the relation is established, allowing a purely x, y based equation to emerge.

Exam trap

A very common mistake is to differentiate dy/dx (in terms of t) directly with respect to t and call it $\frac{d^2y}{dx^2}$ — this is WRONG; one must differentiate dy/dx with respect to x using $\frac{dt}{dx} = \frac{1}{dx/dt}$ , or equivalently differentiate the relation $\sin t\cdot y_1=m\sin(mt)$ implicitly with respect to x.
Students also often forget to convert cos(mt) back to y and cos t back to x at the final step, leaving the "proved" equation partly in terms of t, which loses marks.

Real-world relevance

This differential equation is a special case of the Chebyshev differential equation, whose solutions (Chebyshev polynomials $T_m(x)=\cos(m\cdot\arccos x)$ ) are used extensively in numerical analysis, polynomial approximation, and signal processing (Chebyshev filters in electronics).

Common Mistakes

1Differentiating $\frac{dy}{dx}$ with respect to t and mislabelling the result as $\frac{d^2y}{dx^2}$ instead of dividing again by $\frac{dx}{dt}$ or differentiating implicitly w.r.t. x.
2Sign errors when substituting $dt/dx = -1/\sin t$ into the chain rule expansion.
3Leaving $\cos(mt)$ instead of substituting back $y = \cos(mt)$ in the final line, so the 'proof' isn't expressed purely in x and y.

Interesting Facts

This equation is a disguised form of the Chebyshev differential equation $(1-x^2)y'' - xy' + n^2y = 0$ , whose polynomial solutions $T_n(x) = \cos(n\cdot\arccos x)$ are named after Russian mathematician Pafnuty Chebyshev (1821-1894).

Chebyshev polynomials minimise the maximum error in polynomial approximation, a property exploited in modern calculators and computer software to approximate functions like sin, cos, and log efficiently.

Chebyshev filters, based on these same polynomials, are a standard building block in analog and digital electronic filter design taught in signal-processing courses.

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Frequently Asked Questions

Why can't we directly write $\frac{d^2y}{dx^2} = d/dt(dy/dx)$ ?

Because $\frac{d^2y}{dx^2}$ means differentiating $\frac{dy}{dx}$ with respect to x, not t. One must either use $d/dt(dy/dx)$ divided by $dx/dt$ again, or differentiate implicitly w.r.t. x using $dt/dx$ .

What identity converts $\sin^2t$ into a function of x?

Since $x=\cos t$ , $\sin^2t = 1-\cos^2t = 1-x^2$ , which is exactly the coefficient appearing in the required equation.

What is the significance of this differential equation?

It is a form of the Chebyshev differential equation, whose solutions are Chebyshev polynomials, widely used in approximation theory and filter design.