UPSC NDA I 2023 Mathematics Solved Paper

Concept:The differential equation is factored by taking $\frac{dy}{dx}$ common to find two possible solutions.Explanation:Start with $\left(\frac{dy}{dx}\right)^{2} - x\frac{dy}{dx} = 0$.Factor: $\frac{dy}{dx}\left(\frac{dy}{dx} - x\right) = 0$.Thus either $\frac{dy}{dx} = 0$ or $\frac{dy}{dx} - x = 0$.From $\frac{dy}{dx} = 0$, integrate to get $y = c$.From $\frac{dy}{dx} = x$, integrate to get $y = \frac{x^{2}}{2} + c$.The general solution is $y = \frac{x^{2}}{2} + C$ (when $C$ includes constant from both cases, as $y=c$ is a special case when $c = \frac{x^{2}}{2}+C$ fails for variable $x$, but actually the family $y=\frac{x^{2}}{2}+C$ covers all solutions; the constant $C$ can be any real number).Now check each option:Option A: $y=2x$ gives $\frac{dy}{dx}=2$, then $\left(\frac{dy}{dx}\right)^{2} - x\frac{dy}{dx}=4-2x$, not zero for all $x$.Option B: $y=2x+4$ gives same derivative, not zero.Option C: $y=x^{2}-1$ gives $\frac{dy}{dx}=2x$, then $4x^{2} - 2x^{2}=2x^{2}$, not zero.Option D: $y=\frac{x^{2}-2}{2} = \frac{x^{2}}{2} - 1$ gives $\frac{dy}{dx}=x$, then $x^{2} - x \cdot x = 0$. It satisfies the equation.Thus option D is correct.Answer:$y=\frac{x^{2}-2}{2}$ is a solution to the given differential equation. The correct option is D (or 4).