CBSE Class 12 Mathematics 2026: question 34

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Step-by-step solution

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1. Differentiate both equations w.r.t. $t$:
   $\frac{dx}{dt}=-\sin t$ and $\frac{dy}{dt}=-m\sin mt$.

   1 mark

2. Divide:
   $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{m\sin mt}{\sin t}$, so $\sin t\frac{dy}{dx}=m\sin mt$.

   1 mark

3. Differentiate again w.r.t. $x$ using the chain rule:
   $\sin t\frac{d^{2}y}{d{x}^2}+\cos t\frac{dt}{dx}\frac{dy}{dx}=m^2\cos mt\frac{dt}{dx}$.

   Multiply by $\sin t$ and use $\sin t\frac{dt}{dx}=\frac{\sin t}{-\sin t}=-1$:
   ${\sin }^{2}t\frac{d^{2}y}{d{x}^2}-\cos t\frac{dy}{dx}=-m^2\cos mt$.

   2 marks

4. Substitute ${\sin }^{2}t=1-{\cos }^{2}t=1-x^2$, $\cos t=x$ and $\cos mt=y$:
   $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+m^{2}y=0$. Hence proved.

   1 mark

Answer: $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+m^{2}y=0$ (proved)

Revise the chapter

• Chapter: Continuity and Differentiability (notes, formulas, where marks are lost and every past question)
• Topic: Parametric and Second Order Derivatives