= Implicit solution
{title2=$F(x,y)=C$}
An <implicit solution> describes a <solution of a differential equation> through a relation $F(x,y)=C$, instead of isolating $y$ in an explicit formula. The relation describes a <level set>; an actual scalar <solution of a differential equation> must be a <differentiable function> $y(x)$ on an <interval> whose <function graph> lies in the equation's domain. A whole <level set> need not be one such <function graph>.
For example, consider
$$
y'=-\frac{x}{y},\qquad D=\{(x,y):y\ne0\}.
$$
<Implicit differentiation> of $x^2+y^2=C$ gives $2x+2yy'=0$. For $C>0$, the <circle> contains two <solutions of a differential equation>,
$$
y_+(x)=\sqrt{C-x^2},\qquad y_-(x)=-\sqrt{C-x^2},\qquad -\sqrt C<x<\sqrt C.
$$
Both satisfy the equation on the stated <open interval>, and their <function graphs> stay in $D$. An <initial condition> $(x_0,y_0)\in D$ fixes $C=x_0^2+y_0^2$ and selects the sign of $y_0$. In particular, $y(0)=2$ gives $C=4$ and the upper branch on $(-2,2)$. The constant alone does not select the branch.<Notes on Diffy Qs: Differential Equations for Engineers by Jiří Lebl/1.3 Separable equations>[§1.3.2 Implicit solutions: implicit differentiation, multiple branches, and selection by an initial condition]{ref}
\Image[/implicit-solution-circle-local-graphs.png]
{title=A circular level set gives local solution graphs away from vertical tangencies, while a singular level needs separate analysis}
The precise local criterion uses the <implicit function theorem>. Let $F$ be <continuously differentiable> on an <open subset> of the plane, with $F(x_0,y_0)=C$. If its <partial derivative> $F_y(x_0,y_0)\ne0$, the nearby <level set> is a unique <continuously differentiable> <function graph> $y=g(x)$ through that point. The <chain rule> gives
$$
F_x(x,g(x))+F_y(x,g(x))g'(x)=0,\qquad g'(x)=-\frac{F_x(x,g(x))}{F_y(x,g(x))}.
$$
This proves that the local <function graph> solves $y'=-F_x/F_y$ there. For a different prescribed equation $y'=f(x,y)$, one must also check $F_x+F_yf=0$ along that branch and that it stays in the domain of $f$. A <level set> by itself does not verify an arbitrary <ordinary differential equation>.<Basic Analysis: Introduction to Real Analysis, Volume II by Jiří Lebl/8.5 Inverse and implicit function theorems>[§8.5.1, Theorem 8.5.6 and the preceding circle example: local existence, uniqueness and derivative of an implicit graph]{ref}
For $F=x^2+y^2$, the upper branch through $(1,\sqrt3)$ has $F_y=2\sqrt3$ and <tangent line> <slope> $-1/\sqrt3$. At $(2,0)$, however, $F_y=0$ and $F_x=4$. The <circle> remains a smooth <regular level set> with a vertical <tangent line>: it can be written locally as $x=\sqrt{4-y^2}$, but not as a single <function graph> $y=g(x)$ over an <open interval> containing $2$. Moreover, $-x/y$ is undefined there, so the scalar <solution of a differential equation> cannot include this point in its domain. Changing the parameter draws the curve through the point but does not extend that scalar <initial value problem>.
When both <partial derivatives> vanish, even the geometry needs separate inspection. The level $x^2+y^2=0$ is an isolated point, not a <function graph> on any nonempty <open interval>. The zero <level set> of $G(x,y)=y^2-x^2$ instead consists of the two <lines> $y=x$ and $y=-x$ crossing at the origin, where $\nabla G=0$. Each <line> satisfies the implicit <ordinary differential equation> $yy'=x$, including at the origin; the explicit form $y'=x/y$ excludes the origin. Thus algebraically dividing by a vanishing coefficient changes the domain and can change which <solutions of a differential equation> are admissible. Vanishing $F_y$ alone is only failure of a sufficient test: $F(x,y)=y^3$ has the smooth zero <function graph> $y=0$ despite $F_y=0$ everywhere on it.
<Implicit solutions> are especially useful for <exact first-order ordinary differential equations> and for finding a <first integral of an ordinary differential equation>: a conserved relation can reveal allowed states without an explicit formula. For the normalized <harmonic oscillator> written as $\dot x=y$, $\dot y=-x$, the <energy> $H=(x^2+y^2)/2$ has $\dot H=x\dot x+y\dot y=0$. Its positive <level sets> are the same <circles>. Where $y\ne0$, eliminating time gives $dy/dx=-x/y$ and the local <implicit solution> above. The <harmonic oscillator>'s <parametric curve> continues through $y=0$ because its <vector field> governing evolution in time is defined there; treating $x$ as the independent variable fails at that turn. The origin is an <equilibrium point> of the <harmonic oscillator>, although it is excluded from the scalar equation $y'=-x/y$.
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