General solution of the equation $y=x\dfrac{dy}{dx}+\dfrac {dx}{dy}$ represents _____________.

$\begin{array}{l} y=x \frac{d y}{d x}+\frac{d x}{d y} \\ \Rightarrow y=x\left(\frac{d y}{d x}\right)+\frac{1}{(d y / d x)} \\ Let,\frac{d y}{d x}=p \ldots \text { (1) } \\ y=p x+\frac{1}{p} \\ \Rightarrow p y=p^{2} x+1 . \end{array}$

Diff. w.r.t $x$

$\begin{aligned} y \cdot \frac{d p}{d x}+p(d y / d x)=2 p \cdot x \frac{d p}{d x}+p^{2} . \\ \Rightarrow y \cdot \frac{d p}{d x}+p^{2}=2 p x \frac{d p}{d x}+p^{2} \\ \Rightarrow 2 p x \frac{d p}{d x}-y \cdot \frac{d p}{d x}=0 \\ \Rightarrow \frac{d p}{d x}(2 p x-y)=0 \\ \quad p=y / 2 x \\ \Rightarrow \frac{d y}{d x}=\frac{y}{2 x} \\ \Rightarrow \int \frac{d y}{y}=\frac{1}{2} \int \frac{d x}{x} \\ \Rightarrow \ln y=\frac{1}{2} \ln x+\ln k \\ \Rightarrow \ln y^{2}=\ln x+\ln c \\ \Rightarrow \ln y^{2}=\ln (x c) \\ \Rightarrow y^{2}=c x \end{aligned}$

$\therefore$ Hence the correct option b.