Let $y(x)$ be a solution of $x d y+y d x+y^2(x d y-y d x)=0$ satisfying $y(1)=1$.
Statement-1: The range of $y(x)$ has exactly two points.
Statement-2: The constant of integration is zero.
Answer & explanation
Correct answer: option 2
We have,
$x d y+y d x+y^2(x d y-y d x)=0$
$\Rightarrow d(x y)+x^2 y^2\left\{\frac{x d y-y d x}{x^2}\right\}=0$
$\Rightarrow \frac{1}{(x y)^2} d(x y)+d\left(\frac{y}{x}\right)=0$
On integrating, we get
$-\frac{1}{x y}+\frac{y}{x}=C$ ....(i)
It is given that $y=1$ when $x=1$
∴ $-1+1=C \Rightarrow C=0$
Putting $C=0$ in (i), we get
$-\frac{1}{x y}+\frac{y}{x}=0 \Rightarrow y^2-1=0 \Rightarrow y= \pm 1$
Clearly, statement-1 and statement-2 are true.
Also, statement- 2 is not a correct explanation for statement- 1 .