The differential equation obtained from the function \(y=a(x-a)^{2}\) is

(B)
\(y =a(x-a)^{2}...(1) \)
\( \therefore \frac{d y}{d x} =2 a(x-a)...(2)\)
Eq. \((2)+\mathrm{cq} \cdot(1)\) gives
\(\frac{\left(\frac{d y}{d x}\right)}{y}=\frac{2 a(x-a)}{a(x-a)^{2}} \Rightarrow(x-a)=\frac{2 y}{\left(\frac{d y}{d x}\right)} \text { and } a=x-\frac{2 y}{\left(\frac{d y}{d x}\right)}\)
Substituting these values of \(a\) and \((x-a)\) in eq. (1), we get
\(y =\left[x-\frac{2 y}{\left(\frac{d y}{d x}\right)}\right]\left[\frac{2 y}{\left(\frac{d y}{d x}\right)}\right]^{2} \)
\( =\left[\frac{x\left(\frac{d y}{d x}\right)-2 y}{\left(\frac{d y}{d x}\right)}\right]\left[\frac{4 y^{2}}{\left(\frac{d y}{d x}\right)^{2}}\right] \)
\( \therefore \frac{(y)\left(\frac{d y}{d x}\right)^{3}}{x\left(\frac{d y}{d x}-2 y\right)}=4 y^{2}\)