Q245 / 270 · NDA 2016 (Phase 2) · Mathematics

If $x d y=y d x+y^2 d y, y \gt 0$ and $y(1)=1$, then what is $y(-3)$ equal to?

  • Correct answer
    A
    3 only
  • B
    -1 only
  • C
    Both -1 and 3
  • D
    Neither-1 nor 3

Solution

$\begin{aligned} & \text { Given, } x d y=y d x+y^2 d y \\ & \Rightarrow 1=\frac{y}{x} \frac{d x}{d y}+\frac{y^2}{x} \\ & \Rightarrow \frac{d x}{d y}+y=\frac{x}{y}\end{aligned}$ $\Rightarrow \frac{d x}{d y}-\frac{x}{y}=-y$ ...(1) $\begin{aligned} & P=-\frac{1}{y}, Q=-y \\ & \mathrm{IF}=e^{\int P d y}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=\frac{1}{y} \end{aligned}$ Now, Solution of d.E. $\begin{aligned} & \mathrm{x}(\mathrm{I} . \mathrm{F})=\int(\mathrm{Q} . \mathrm{I} . \mathrm{F} .) \mathrm{dy} \\ & \frac{x}{y}=\int \frac{1}{y}(-y) d y+C \\ & \Rightarrow \frac{x}{y}=\int-1 d y+C \\ & \Rightarrow \frac{x}{y}=-y+C \\ & y(1)=1 \end{aligned}$ $\begin{aligned} & \frac{1}{1}=-1+C \Rightarrow C=2 \\ & \Rightarrow \frac{x}{y}=-y+2 \Rightarrow x=-y^2+2 y \\ & \Rightarrow y(-3) \Rightarrow-3=-y^2+2 y \\ & \Rightarrow y^2-2 y-3=0 \\ & \Rightarrow y=\frac{+2 \pm \sqrt{4+12}}{2}=\frac{2 \pm 4}{2} \\ & \Rightarrow y=3,-1 \\ & \text {Since } y \gt 0 \text { so } y=3\end{aligned}$

Hence, the correct answer is option: 'A'.

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