UPSC NDA I 2019 Mathematics Solved Paper

Concept:Substitution method reduces the differential equation to a separable form.Integration of $\sec t$ is used.Trigonometric identity simplifies to the required form.Explanation:Given: $\frac{dy}{dx} = \cos(y - x) + 1$.Let $t = y - x$.Differentiating: $\frac{dy}{dx} = 1 + \frac{dt}{dx}$.Substitute: $1 + \frac{dt}{dx} = \cos t + 1$.Hence $\frac{dt}{dx} = \cos t$.Separate variables: $\sec t \, dt = dx$.Integrate both sides: $\int \sec t \, dt = \int dx$.We know $\int \sec t \, dt = \ln|\sec t + \tan t| + C$.Thus $\ln|\sec t + \tan t| = x + C$.Take antilog: $\sec t + \tan t = e^x \cdot e^C = k e^x$, where $k = e^C$.Put back $t = y - x$: $\sec(y - x) + \tan(y - x) = k e^x$.Using identity $\sec^2 A - \tan^2 A = 1$, we get $(\sec A + \tan A)(\sec A - \tan A) = 1$.So $\sec A + \tan A = \frac{1}{\sec A - \tan A}$.Therefore $\frac{1}{\sec(y - x) - \tan(y - x)} = k e^x$.Rearrange: $e^x [\sec(y - x) - \tan(y - x)] = \frac{1}{k} = c$, a constant.This matches option A.Answer:Option A: $e^{x} [\sec (y - x) - \tan (y - x)] = c$