UPSC NDA I 2021 Mathematics Solved Paper

Concept:The order of a differential equation is the highest order of derivative present in the equation after eliminating all arbitrary constants.Explanation:The given general solution is $y^{2} + 2cy - cx + c^{2} = 0$, where $c$ is an arbitrary constant.Rewrite as $(y + c)^{2} = cx$.Differentiate both sides with respect to $x$: $2(y + c)y' = c$.Solve for $c$: $c = \frac{2yy'}{1 - 2y'}$.Substitute this $c$ back into the original equation to eliminate the constant:$(y + \frac{2yy'}{1 - 2y'})^{2} = \frac{2yy'}{1 - 2y'} \cdot x$.Simplify step by step: $y^{2} = 2xyy'(1 - 2y')$.Expand: $y^{2} - 2xyy' + 4xy(y')^{2} = 0$.The highest derivative appearing in this equation is $y'$ (the first derivative).Therefore, the order of the differential equation is 1.(Note: The degree of the equation is 2, but the question asks only for order.)Answer:The order of the differential equation is 1.