Question 345 marksLong AnswerSection D

If x=cos⁡t,y=cos⁡mtx = \cos t, y = \cos mt, prove that (1−x2)d2ydx2−xdydx+m2y=0(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + m^2 y = 0.

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  ⟹  dx/dt=−sin⁡tx = \cos t \implies dx/dt = -\sin t.
  • y=cos⁡mt  ⟹  dy/dt=−msin⁡mty = \cos mt \implies dy/dt = -m\sin mt.
  • dydx=dy/dtdx/dt=msin⁡(mt)/sin⁡t\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = m\sin(mt)/\sin t, so sin⁡t⋅(dy/dx)=msin⁡(mt)\sin t \cdot (dy/dx) = m\sin(mt).

Differentiating again

  • Differentiating sin⁡t⋅y1=msin⁡(mt)\sin t\cdot y_1 = m\sin(mt) with respect to x (using dt/dx=−1/sin⁡tdt/dx = -1/\sin t) gives, after simplification, −x⋅y1+(1−x2)⋅y2=−m2y-x\cdot y_1 + (1-x^2)\cdot y_2 = -m^2 y (where y1=dydxy_1=\frac{dy}{dx}, y2=d2ydx2y_2=\frac{d^2y}{dx^2}).

Conclusion

  • Rearranging gives exactly (1−x2)y2−xy1+m2y=0(1-x^2)y_2 - x y_1 + m^2y = 0, which is the required result — hence proved.
Answer keywords:parametric differentiationsecond order derivativechain ruleChebyshev differential equationimplicit differentiationd2y/dx2

Marking scheme

  1. 1.1 mark: correctly finding dx/dt=−sin⁡tdx/dt = -\sin t and dy/dt=−msin⁡(mt)dy/dt = -m\sin(mt).
  2. 2.1 mark: correctly forming dydx=msin⁡(mt)/sin⁡t\frac{dy}{dx} = m\sin(mt)/\sin t, i.e. sin⁡t⋅y1=msin⁡(mt)\sin t \cdot y_1 = m\sin(mt).
  3. 3.2 marks: correctly differentiating implicitly w.r.t. x (using dt/dx=−1/sin⁡tdt/dx = -1/\sin t) to obtain −xy1+(1−x2)y2-x y_1 + (1-x^2) y_2 on the LHS.
  4. 4.1 mark: correctly simplifying the RHS to −m2y-m^2y (using cos⁡(mt)=y\cos(mt) = y) and concluding (1−x2)y2−xy1+m2y=0(1-x^2)y_2 - xy_1 + m^2y = 0.

Hint

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

Quick oral answer

Since x=cos⁡tx=\cos t and y=cos⁡mty=\cos mt, dydx=msin⁡(mt)/sin⁡t\frac{dy}{dx} = m\sin(mt)/\sin t; differentiating this again with respect to x and simplifying using sin⁡2t=1−x2\sin^2t=1-x^2 and cos⁡(mt)=y\cos(mt)=y gives exactly (1−x2)y′′−xy′+m2y=0(1-x^2)y''-xy'+m^2y=0.

Common mistakes

  1. 1.Differentiating dydx\frac{dy}{dx} with respect to t and mislabelling the result as d2ydx2\frac{d^2y}{dx^2} instead of dividing again by dxdt\frac{dx}{dt} or differentiating implicitly w.r.t. x.
  2. 2.Sign errors when substituting dt/dx=−1/sin⁡tdt/dx = -1/\sin t into the chain rule expansion.
  3. 3.Leaving cos⁡(mt)\cos(mt) instead of substituting back y=cos⁡(mt)y = \cos(mt) in the final line, so the 'proof' isn't expressed purely in x and y.

Analysis and 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, dydx\frac{dy}{dx} is found as (dy/dt)÷(dx/dt)(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−x2\sin^2t = 1-\cos^2t = 1-x^2, so 1−x2\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 d2ydx2\frac{d^2y}{dx^2} — this is WRONG; one must differentiate dy/dx with respect to x using dtdx=1dx/dt\frac{dt}{dx} = \frac{1}{dx/dt}, or equivalently differentiate the relation sin⁡t⋅y1=msin⁡(mt)\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 Tm(x)=cos⁡(m⋅arccos⁡x)T_m(x)=\cos(m\cdot\arccos x)) are used extensively in numerical analysis, polynomial approximation, and signal processing (Chebyshev filters in electronics).

Interesting facts

This equation is a disguised form of the Chebyshev differential equation (1−x2)y′′−xy′+n2y=0(1-x^2)y'' - xy' + n^2y = 0, whose polynomial solutions Tn(x)=cos⁡(n⋅arccos⁡x)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.

Question details

Q34 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section D
Marks
5
Type
Long Answer
Chapter
Continuity and Differentiability
Topic
Second Order Derivatives of Parametric Functions
Bloom's level
analyze
Syllabus topic
Unit 2: Continuity and Differentiability — Second Order Derivative; Parametric Differentiation; Chebyshev-type Differential Equation
Author
CBSE
Reviewed by
Padhantu Academic Team

Spotted a mistake or something unclear?

Tell us and we will fix it.

Frequently asked questions

Why can't we directly write d2ydx2=d/dt(dy/dx)\frac{d^2y}{dx^2} = d/dt(dy/dx)?

Because d2ydx2\frac{d^2y}{dx^2} means differentiating dydx\frac{dy}{dx} with respect to x, not t. One must either use d/dt(dy/dx)d/dt(dy/dx) divided by dx/dtdx/dt again, or differentiate implicitly w.r.t. x using dt/dxdt/dx.

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

Since x=cos⁡tx=\cos t, sin⁡2t=1−cos⁡2t=1−x2\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.