Question

The solution of $(x+logy)dy+ydx=0$ where $y\left(0\right)=1$ is

• A. $y(x-(A\left)\right)+ylogy=0$
• B. $y(x-1+logy)+1=0$
• C. $xy+ylogy+1=0$
• D. None of these

Answer 1

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

$(x+logy)dy+ydx=0$ $\therefore d\left(xy\right)=xdy+ydx$
$(xdy+ydx)+\left(logydy\right)=0$
$xy+(logy\cdot -\frac{y}{y}dy)=0$
$xy+ylogy-y+c=0$
$y(x-1)+ylogy+c=0$
$y(x-1+logy)+c=0$
$1(0-1+log1)+c=0$
$c=1$
given $y\left(0\right)=1$
i.e. (0, 1) is the solution of the differential equation
So, overall solutions is
$y(x-1+logy)+1=0$