Implicit solution to ODE

An implicit solution to an ODE is one where $x$ and $y$ are related by an equation that doesn’t isolate $y$ on one side. Given an implicit relation, you can verify whether it satisfies an ODE — and apply initial conditions to pick out a specific branch.

Example

Consider the implicit relation $x^2+y^2=1$ on $x\in (-1,1)$. Is $y(x)$ a solution to the ODE $y^{\prime }=-\frac{x}{y}$?

Step 1: Differentiate implicitly with respect to $x$.

$2x+2y{y}^{\prime }=0$

Step 2: Solve for $y^{\prime }$.

$y^{\prime }=-\frac{x}{y}$

This matches the original ODE exactly. So any function $y(x)$ implicitly defined by $x^2+y^2=1$ satisfies the ODE.

Step 3: Solve for $y$ explicitly.

$y(x)=\pm \sqrt{1-x^2}$

Two branches — upper semicircle ($+$) and lower semicircle ($-$).

Step 4: Verify both candidates.

For $y=+\sqrt{1-x^2}$: $y^{\prime }=\frac{-x}{\sqrt{1-x^2}}$. Check: $-\frac{x}{y}=\frac{-x}{\sqrt{1-x^2}}$. ✓

For $y=-\sqrt{1-x^2}$: $y^{\prime }=\frac{x}{\sqrt{1-x^2}}$. Check: $-\frac{x}{y}=\frac{-x}{-\sqrt{1-x^2}}=\frac{x}{\sqrt{1-x^2}}$. ✓

Both branches satisfy the ODE.

Step 5: Apply initial conditions to pick a branch.

If $y(0)=1$: then $y_2(x)=\sqrt{1-x^2}$ gives $y_2(0)=1$ ✓. So this is the unique solution.

If $y(0)=-1$: then $y_1(x)=-\sqrt{1-x^2}$ would be the answer.

Why implicit solutions matter

Many ODEs don’t have nice explicit solutions. Examples:

• $y{e}^{2y}{y}^{\prime }=x{e}^{-x}$ separates and integrates to $e^{2y}(2y-1)=-4e^{-x}(x+1)+c$ — implicit.
• An Exact equation’s solution is given by $F(x,y)=c$ where $F$ is the potential function — almost always implicit.

In these cases the relation $F(x,y)=c$ tells you everything about the solution curves, just not in $y=f(x)$ form.

Verifying implicit solutions

The general procedure:

1. Differentiate the relation with respect to $x$, treating $y$ as a function of $x$.
2. Solve for $y^{\prime }$ to get the ODE the relation satisfies.
3. Compare with the given ODE.
4. Apply initial conditions if needed to pick out a specific branch (when the implicit relation produces multiple $y$-values for the same $x$).

The implicit form is sometimes the best form. Trying to convert to explicit can introduce branch ambiguities or lose mathematical content.