Question

The general solution of the D.E $y(1+\log x)dx-x\log xdy=0$ is ?

• A. $x\log x=cy$
• B. $y\log y=cx$
• C. $xy=c$
• D. $\frac{x}{y}=c$

Answer 1

The correct option is A $x\log x=cy$

$y(1+\log x)dx-x\log xdy=0\Rightarrow y(1+\log x)dx=x\log xdy\Rightarrow y(1+\log x)=x\log x\frac{dy}{dx}\Rightarrow \frac{y(1+\log x)}{x\log x}=\frac{dy}{dx}\Rightarrow \frac{dy}{y}=(\frac{1}{x\log x}+\frac{1}{x})dx\Rightarrow \int \frac{dy}{y}=\int (\frac{1}{x\log x}+\frac{1}{x})dx(\int \frac{1}{x}dx=\log x)\Rightarrow \log y+\log C=\log \left(\log x\right)+\log x\Rightarrow \log Cy=\log \left(x\log x\right)(\log a+\log b=\log ab)\Rightarrow x\log x=Cy$