 Solution

For the given differential form, set $P=x(y^4+1)$ and $Q=y(2x^2{y}^2+1)$. Their partial derivatives satisfyIn fact an explicit potential issince $F_x=P$ and $F_y=Q$. Thus the differential form is an exact differential form on all of ${\mathbb{R}}^2$.

Replacing $y^2$ by $u$ suggests the potentialDirect differentiation verifies that $dH/dx$ is exactly the left-hand side of the required ordinary differential equation. This direct verification remains valid even for negative $u$; it does not require a real square root $y=\sqrt{u}$. ThereforeThe initial condition gives $C=1$. Solving this quadratic equation in $u$ and selecting the branch that has $u(1)=0$ givesThe second expression also defines the removable value $u(0)=1$. The maximal open interval containing $1$ on which this is a differentiable real solution isAt either endpoint the square root vanishes and $2x^{2}u+1=0$, while $x(u^2+1)\ne 0$, so the original ordinary differential equation cannot hold there with a finite derivative.