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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, considerImplicit differentiation of $x^2+y^2=C$ gives $2x+2y{y}^{\prime }=0$. For $C>0$, the circle contains two solutions of a differential equation,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.^[ref]

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 $F_y(x_0,y_0)\ne 0$, the nearby level set is a unique continuously differentiable function graph $y=g(x)$ through that point. The chain rule givesThis proves that the local function graph solves $y^{\prime }=-F_x/F_y$ there. For a different prescribed equation $y^{\prime }=f(x,y)$, one must also check $F_x+F_{y}f=0$ 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.^[ref]

For $F=x^2+y^2$, the upper branch through $(1,\sqrt{3})$ has $F_y=2\sqrt{3}$ and tangent line slope $-1/\sqrt{3}$. 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 $y{y}^{\prime }=x$, including at the origin; the explicit form $y^{\prime }=x/y$ 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 $y\ne 0$, 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 $y^{\prime }=-x/y$.

Where $y\ne 0$ and $x^2\ne 1$, the separable differential equation givesThe requested point selects the implicit solution $2y^2+1=(x^2-1{)}^2$. The two real branches approaching that point areThere is an important initial value problem qualification: the differential equation as printed has no classical solution through the specified initial point. Its right side is undefined at $y=0$, and these branches have an infinite one-sided derivative as $x\downarrow \sqrt{2}$. They satisfy the differential equation on $x>\sqrt{2}$ and extend continuously to the requested point. Multiplying by the vanishing denominator produces an implicit differential equation with a larger domain; it does not repair the original initial value problem.

For continuously differentiable coefficients on a simply connected domain, the exactness condition is $P_y=Q_x$. Locally the same condition suffices without a global topological assumption. An exact first-order ordinary differential equation has a potential $F$ with $F_x=P$ and $F_y=Q$, so $dF/dx=P+Q{y}^{\prime }=0$ along a solution. Thus its implicit solution is $F(x,y)=C$. One local construction, on a rectangle about $(x_0,y_0)$, isThe equality of cross derivatives verifies both required derivatives; on a simply connected domain one may use the corresponding path-independent integral of the exact differential form.

Here $P=4x+3y$, $Q=3x+3y^2$ and $P_y=Q_x=3$. Integrating $P$ with respect to $x$ gives $F=2x^2+3xy+K(y)$. Matching $F_y=Q$ gives $K^{\prime }=3y^2$, so take $K=y^3$. The initial value sets $C=F(1,2)=16$. ThereforeThis is the requested explicit relation between the coordinates. Since $F_y(1,2)=15\ne 0$, the implicit function theorem gives a unique local graph through the initial point, with slope $-10/15=-2/3$.

For this homogeneous first-order differential equation, set $v=y/x$, with $x\ne 0$. The product rule gives $y^{\prime }=v+x{v}^{\prime }$, and substitution simplifies the equation toThe useful further substitution is $w=e^{-v}-v^2$. Its chain rule derivative satisfiesThus $(x^{2}w{)}^{\prime }=0$. Replacing $v$ and $w$ gives the implicit solutionNo division by $w$ was needed, so the level $C=0$ is included. A direct verification also identifies the integrating factor for a differential one-form:Setting this exact differential to zero recovers precisely the original equation after multiplication by the nonzero factor $e^{y/x}$. The coefficient $1+2v{e}^v$ never vanishes for real $v$, since $v{e}^v\ge -1/e$ implies $1+2v{e}^v\ge 1-2/e>0$. Hence these level sets give local function graphs $y(x)$ everywhere with $x\ne 0$.

A first-order ordinary differential equation is separable if its explicit form factors as $y^{\prime }=a(t)b(y)$. On an interval where $b(y)\ne 0$, write $dy/b(y)=a(t)dt$ and take antiderivatives to obtain an implicit solution:An initial condition determines $C$ and selects a branch and domain. The useful simplification is that the two integrals each involve only one variable, even when the original equation is nonlinear.

For Newton's law of cooling, division by $y-T_a$ gives $dy/(y-T_a)=-kdt$. To keep the logarithm dimensionless, integrate relative to the initial excess temperature:When $y_0\ne T_a$, continuity preserves the sign of the excess temperature until any potential zero, giving $y-T_a=(y_0-T_a)e^{-k(t-t_0)}$. This formula never reaches zero at finite time. Taking its derivative verifies the ordinary differential equation, and substitution at $t_0$ verifies the initial condition.

Always check what division removed. Each zero $b(y_{\ast })=0$ gives a constant solution of a differential equation $y=y_{\ast }$. Cooling therefore also has the equilibrium point $y=T_a$, omitted by division but included in the final explicit formula. For the nonunique initial value problem $y^{\prime }=\sqrt{|y|}$, division by $\sqrt{|y|}$ only describes the moving pieces: a valid solution may remain zero for an arbitrary time before joining one. Separation is a method for finding solutions, not a uniqueness theorem.