v(v-2p)+(vp-p^2)/p'=0
Since this is an homogeneous ODE and p(0)=0 then p=kv
v(v-2kv)+(v(kv)-(kv)^2)/k=0
(v^2)(2-3k)=0
k=2/3
The solution is : p=2v/3
.
For information, the complete resolution is:
v(v-2p)p'+(v-p)p=0
p=vt
p'=t+vt'
v(v-2vt)(t+vt')+(v-vt)vt=0
(1-2t)(t+vt')+(1-t)t=0
(1-2t)(t')+t(2-3t)/v=0
((2t-1)/(t(3t-2)))t'=-1/v
(1/2)ln(t)+(1/6)ln(3t-2)=-ln(v)+c
(3t-2)(t^3)=C/(v^6)
(3(p/v)-2)(p^3)/(v^3)=C/(v^6)
(3p-2v)(p^3)=C/(v^2)
3(p^4)-2v(p^3)-C/v^2=0
p=p(v) can be obtained by solving this quartic equation (unknown p) using the arduous Ferrari's formulae.
In particular case C=0, the equation reduces to (p^3)(3p-2v)=0 leading to the obvious particular solution p=2v/3.