Question

The solution to the differential equation $y\ln y+x{y}^{\prime }=0$ where$y\left(1\right)=e$, is:

• A. $x\left(\ln y\right)=1$
• B. $xy\left(\ln y\right)=1$
• C. $(\ln y{)}^2=2$
• D. $\ln y+\left(\frac{x^2}{2}\right)y=1$

Answer 1

The correct option is A $x\left(\ln y\right)=1$

$y\ln y+x{y}^{\prime }=0\Rightarrow \frac{dy}{dx}=-\frac{\left(\log y\right)y}{x}\Rightarrow \frac{dy}{\left(\log y\right)y}=-\frac{1}{x}$
Integrating both sides w.r.t x
$\int \frac{dy}{\left(\log y\right)y}=\int -\frac{1}{x}dx\Rightarrow \log \log y=-\log x+c\Rightarrow y=e^{c/x}$
Now $y\left(1\right)=e\Rightarrow x\ln y=1$