Explanation:
(C) : The given differential equation is, \(y^{2}=2 c\left(x+c^{2 / 3}\right)\)
On differentiating both side, we get-
\(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{c}(1+0)\)
\(\mathrm{c}=\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}\)
Putting the value of \(\mathrm{c}\) in equation (i), we get-
\(y^{2}=2 y \frac{d y}{d x}\left[x+\left(y \frac{d y}{d x}\right)^{2 / 3}\right] \Rightarrow y^{2}-2 x y \frac{d y}{d x}=2\left(y \frac{d y}{d x}\right)^{\frac{2}{3}+1}\)
\(y^{2}-2 x y \frac{d y}{d x}=2\left(y \frac{d y}{d x}\right)^{5 / 3}\)
Take cube on both side, we get-
\(\left(y^{2}-2 x y \frac{d y}{d x}\right)^{3}=2^{3}\left(y \frac{d y}{d x}\right)^{5}\)
So, Order \(=1\), Degree \(=5\)