VITEEE 2012 Paper

ρ =

[1+( dy dx )2]3∕2

d2y dx2

⇒ ρ (

d2y

dx2

) = [1+(

dy

dx

)2]3∕2
On squaring both sides, we get
ρ2 (

d2y

dx2

)2 = [1+(

dy

dx

)2]3
Clearly, it is a second order differential equation of degree 2.
Note that the higher order derivative is in the transcendental, then we do not determined the degree of that equation