How do I solve this differential equation $y''^2-2y'y''+3=0$ in parametric form? I do have quite no idea about this one.
The obvious substitution $y'=p, y''=p\frac{dp}{dy}$ doesn't make the situation any better

Note that $2y'y'' = ((y')^2)'$
Then let $g = y'$ so the equation reformulates to:
$$(g')^2 - g^2 + 3 =0 $$ which is much simpler an easier to solve. Then find a solution for $g = g(x) = y' $ and integrate to find $y(x) $
