The solution curve y = y(x) of the differential equation ydx – xdy = 0, passing through (2, 4), does not pass through the point

$\left(1, \frac{1}{2}\right)$

$y=y(x)$

so  $\frac{y d x-x d y}{y^2}=\frac{0}{y^2} \Rightarrow d(\frac{x}{y})=0$

as  $d(\frac{x}{y}) = \frac{ydx - xdy}{y^2}$

So integrating both sides

so $\frac{x}{y} = c$  →  passes through (2, 4)

$\Rightarrow \frac{2}{4}=c \Rightarrow c = \frac{1}{2} \Rightarrow \frac{x}{y}=\frac{1}{2}$

as  $\frac{1}{\frac{1}{2}} \neq \frac{1}{2} \Rightarrow \frac{2}{1} \neq \frac{1}{2}$  at  $P(1, \frac{1}{2})$