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

The general solution of the differential equation $y(1+\log x)\left(\frac{d x}{d y}\right)$ $-x \log x=0$ is
  1. $\mathrm{y}(1+\log \mathrm{x})=\mathrm{c}$
  2. $\mathrm{x} \log \mathrm{x}=\mathrm{yc}$
  3. $x \log x=y+c$
  4. $\log \mathrm{x}-\mathrm{y}=\mathrm{c}$

Solution

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

Asked in: MHT CET 2021 (21 Sep Shift 1)

Practice more Differential Equations questions on Aicharya