Q76 / 150 · MHT CET 2022 (10 Aug Shift 2) · Mathematics

The solution of the differential equation $(1+x) y \mathrm{~d} x+(1-y) x \mathrm{~d} y=0$ is

  • A
    $\log x y-x+y=C$
  • B
    $\log \left(\frac{x}{y}\right)-x+y=C$
  • C
    $\log x y-x-y=C$
  • Correct answer
    D
    $\log (x y)+x-y=C$

Solution

$\begin{aligned} & (1+x) y \mathrm{~d} x+(1-y) x \mathrm{~d} y=0 \\ & \Rightarrow \int \frac{1+x}{x} \mathrm{~d} x=\int \frac{y-1}{y} \mathrm{~d} y \\ & \Rightarrow \log |x|+x=y-\log |y|+C \\ & \Rightarrow \log (x y)+x-y=C\end{aligned}$

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

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