Correct option is (b) \(y(p) = p^2 + \frac{2c}{p}\)
The explanation is: Let y’=p –> y = 2xp – 3p^2 ….(1) is in the Lagrange equation form
now differentiating we get dy=2xdp+2pdx-6pdp and dy=pdx
thus pdx=2xdp+2pdx-6pdp→-pdx=2xdp-6pdp –> \(\frac{dx}{dp} + \frac{2}{p} x – 6=0\)…(2)
(2) is a linear D.E whose I.F=\(e^{\int \frac{2}{p} \,dp} = p^2\) hence its solution is
\(p^2 x(p) = \int 6p^2 \,dp + c \rightarrow x(p) = 2p + \frac{c}{p^2}\) ….substituting in (1) we get
\(y(p) = 2(2p+\frac{c}{p^2})p-3p^2 = p^2 + \frac{2c}{p}.\)