The particular solution of the differential equation $y(1+\log x) \frac{d x}{d y}-x \log x=0$ when $x=e,…
The particular solution of the differential equation $y(1+\log x) \frac{d x}{d y}-x \log x=0$ when $x=e, y=e^2$ is
- $y^2=e^4 \log x$
- $y=e^2 \log x$
- $y=x^2 \log x$
- $y=e x \log x$
Solution
$\begin{aligned} & y(1+\log x) \frac{d y}{d x}-x \log x=0 \\ & \therefore y(1+\log x) \frac{d y}{d x}=x \log x \Rightarrow \frac{d y}{d x}=\frac{y(1+\log x)}{x \log x} \\ & \therefore \int \frac{d y}{d x}=\int \frac{1+\log x}{x \log x} d x \\ & \therefore \log |y|+\log |x \log x|+\log c \\ & \text { We have } x=e, y=e^2 \\ & \therefore \log |e|^2=\log |e \log e|+\log c \\ & 2=1+\log c \Rightarrow \log c=1=\log e \\ & \therefore \log |y|=\log |x \log x|+\log e \\ & y=e x \log x\end{aligned}$
Asked in: MHT CET 2021 (24 Sep Shift 1)
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