# Question:why odesdvisor says this ODE is quadrature?

## Question:why odesdvisor says this ODE is quadrature?

Maple 2019

I am not able to understand why this ODE is quadrature. It is first order ODE of second degree. Solving for y'(x) gives two ODE's. Only one of these two ODE's is quadrature and the second is Abel.

So  why and how did odesdvisor come to conclusion that it is  quadrature? Did it pick the first ODE that comes from solving for y'(x)?

Note that from help, quadrature is ODE (for first order) is one which
the ODE is of first order and the right hand sides below depend only on x or y(x)

And the above definition only applied here for one of the 2 ODE's embeded inside this first order ODE of second degree. So I am just trying to understand the logic behind this result of odeadvisor

 > ode:= (x^2-a*y(x))*diff(y(x),x)^2-2*x*y(x)*diff(y(x),x) = 0;

 > odes:=[solve(ode,diff(y(x),x))]; #solve for y' we get 2 first order ODE's

 > DEtools:-odeadvisor(diff(y(x),x)=odes[1]); #find type of first one

 > DEtools:-odeadvisor(diff(y(x),x)=odes[2]); #find type of second one

btw, the above is just one example. I have many more. below show one more such example

 > #example 2
 > ode:=diff(y(x),x)^3-(2*x+y(x)^2)*diff(y(x),x)^2+(x^2-y(x)^2+2*x*y(x)^2)*diff(y(x),x)-(x^2-y(x)^2)*y(x)^2 = 0;

 > odes:=[solve(ode,diff(y(x),x))]; #solve for y' we get 3 first order ODE's

 > DEtools:-odeadvisor(diff(y(x),x)=odes[1]); #find type of first one

 > DEtools:-odeadvisor(diff(y(x),x)=odes[2]); #find type of second one

 > DEtools:-odeadvisor(diff(y(x),x)=odes[3]); #find type of third one

 >