UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The differential equation is manipulated into a form involving $d\left(\frac{x}{y}\right)$, enabling direct integration to find the particular solution.Explanation:Start with $x\,dy = y\,dx + y^2\,dy$.Rearranging gives $x\,dy - y\,dx = y^2\,dy$.Divide both sides by $y^2$: $\frac{x\,dy - y\,dx}{y^2} = dy$.Recall that $d\left(\frac{x}{y}\right) = \frac{y\,dx - x\,dy}{y^2} = -\frac{x\,dy - y\,dx}{y^2}$.Thus $\frac{x\,dy - y\,dx}{y^2} = -d\left(\frac{x}{y}\right)$.Substitute: $-d\left(\frac{x}{y}\right) = dy$.Integrate both sides: $-\frac{x}{y} = y + c$.Apply initial condition $y(1)=1$: $-\frac{1}{1} = 1 + c \Rightarrow c = -2$.Hence $-\frac{x}{y} = y - 2$.Multiply by $y$: $-x = y^2 - 2y$, or $y^2 - 2y + x = 0$.To find $y(-3)$, put $x = -3$: $y^2 - 2y - 3 = 0$.Factor: $(y-3)(y+1)=0$, giving $y = 3$ or $y = -1$.Since $y > 0$, the valid answer is $y = 3$.Answer:$y(-3) = 3$, corresponding to option A (3 only).