Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
If the independent variable $x$ is changed to $y$, then the expression $x\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 - \frac{dy}{dx} = 0$ is transformed to $x\frac{d^2x}{dy^2} + \left(\frac{dx}{dy}\right)^2 = k\frac{dx}{dy}$ then $k$ equals.
Step-by-Step Solution
Key Concept: Convert between $\frac{d^2y}{dx^2}$ and $\frac{d^2x}{dy^2}$ using the chain rule and inverse function relationships.
Given $\frac{d^2y}{dy^2} = -\left(\frac{dx}{dy}\right) \frac{d^2y}{dx^2}$, we derive $\frac{d^2x}{dx^2} = -\frac{d^2x}{dy^2} \cdot \frac{1}{(dx/dy)^3}$. After algebraic manipulation, the equation $x\frac{d^2x}{dy^2} + \left(\frac{dx}{dy}\right)^2 = \frac{dx}{dy}$ can be rewritten to yield $\lambda = 1$.
Correct Answer: 1