= Solution
For the given <differential form>, set $P=x(y^4+1)$ and $Q=y(2x^2y^2+1)$. Their <partial derivatives> satisfy
$$
P_y=4xy^3=Q_x.
$$
In fact an explicit potential is
$$
F(x,y)=\frac12x^2(y^4+1)+\frac12y^2,
$$
since $F_x=P$ 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 potential
$$
H(x,u)=\frac12\left[x^2(u^2+1)+u\right].
$$
Direct <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 $y=\sqrt u$. Therefore
$$
x^2u^2+u+x^2=C.
$$
The <initial condition> gives $C=1$. Solving this <quadratic equation> in $u$ and selecting the branch that has $u(1)=0$ gives
$$
\boxed{u(x)=\frac{-1+\sqrt{1+4x^2-4x^4}}{2x^2}=\frac{2(1-x^2)}{1+\sqrt{1+4x^2-4x^4}}.}
$$
The second expression also defines the removable value $u(0)=1$. The maximal open <interval> containing $1$ on which this is a differentiable real solution is
$$
\boxed{-\sqrt{\frac{1+\sqrt2}{2}}<x<\sqrt{\frac{1+\sqrt2}{2}}.}
$$
At either endpoint the square root vanishes and $2x^2u+1=0$, while $x(u^2+1)\ne0$, so the original <ordinary differential equation> cannot hold there with a finite <derivative>.
Back to article page
