Question

The solution of $(x+\log y)dy+ydx=0$ when $y\left(0\right)=1$ is:

• A. $y(x-1)+y\log y=0$
• B. $y(x-1+\log y)+1=0$
• C. $xy+y\log y+1=0$
• D. none of these

Answer 1

The correct option is B $y(x-1+\log y)+1=0$

Given, $(x+\log y)dy+ydx=0$
$xdy+ydx=-\log ydy$
$d\left(xy\right)=-\log ydy$
Now integrating both sides,
$xy=-\int \log y\cdot 1dy$
Now using by parts for R.H.S.
$-xy=y\log y-\int \frac{1}{y}\cdot ydy=y\log y-y+c$
Now $y\left(0\right)=1\Rightarrow c=1$
Hence required solution is,
$y(x-1+\log y)+1=0$