Q88 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics

The general solution of the differential equation $$ y(1+\log x)\left(\frac{d x}{d y}\right) $$ $-x \log x=0$ is

  • A
    $\mathrm{y}(1+\log \mathrm{x})=\mathrm{c}$
  • Correct answer
    B
    $\mathrm{x} \log \mathrm{x}=\mathrm{yc}$
  • C
    $x \log x=y+c$
  • D
    $\log \mathrm{x}-\mathrm{y}=\mathrm{c}$

Solution

$$ \begin{aligned} & y(1+\log x)\left(\frac{d y}{d x}\right)-x \log x=0 \\ & \therefore \int \frac{(1+\log x)}{x \log x} d x=\int \frac{d y}{y} \\ & \int \frac{d x}{x \log x}+\int \frac{d x}{x}=\int \frac{d y}{y} \\ & \therefore \log (\log x)+\log x=\log y+\log c \\ & \therefore \log [x \log x]=\log (y, c) \Rightarrow x \log x=y c \end{aligned} $$

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

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