Implicit Solution Curves

What an Implicit Solution Says

An equation for a solution does not always give \(y\) directly as a formula in \(x\). It may instead relate \(x\) and \(y\). For example, a family of curves can be described by

Here \(C\) is a constant, so different choices of \(C\) describe different curves. The equation specifies which points \((x,y)\) belong to a particular curve without isolating \(y\). Such a relation is called an implicit solution when its differentiable branches satisfy the differential equation under consideration.

The qualification about branches matters. A relation describes a set of points; a solution to an equation \(y'=f(x,y)\) is a differentiable function \(y(x)\) on an interval. To test a branch, regard \(y\) as depending on \(x\), differentiate the relation using the chain rule, and compare the resulting derivative with the differential equation.

From a Separable Equation to an Implicit Family

Consider the differential equation

The original equation is defined when \(y\ne-1\), and it also admits the solution \(y=0\). On a solution where \(y\ne0\), separation and integration lead to the implicit relation above. In brief, separating gives \((1/y+1)\,dy=2x\,dx\); integrating gives \(\ln|y|+y=x^2+C\).

This derivation involved dividing by \(y\), so the integrated family describes solutions with \(y\ne0\). It does not describe every solution of the original equation. In particular, \(y(x)=0\) is a solution of the original equation: its derivative is zero, and the right-hand side is \(2x(0)/(0+1)=0\). The original equation excludes \(y=-1\), but it does not exclude \(y=0\).

Why Implicit Differentiation Recovers the Equation

Suppose a differentiable branch satisfies \(\ln|y|+y=x^2+C\). The logarithm is defined for \(y\ne0\). Differentiating each side with respect to \(x\), and using the chain rule on terms involving \(y\), gives

Factoring out the derivative gives

When \(y\ne0\) and \(y\ne-1\), the coefficient of \(dy/dx\) is nonzero, so division is permitted. Since \(\frac{1}{y}+1=\frac{y+1}{y}\), rearranging yields

This is exactly the original differential equation. The argument explains why each differentiable branch of the implicit family, wherever \(y\ne0,-1\), satisfies the equation. The condition \(y\ne-1\) is needed both because the original equation is undefined there and because the coefficient being divided by would be zero.

Worked Examples

What the Constant and the Relation Tell You

For a fixed value of \(C\), the relation selects one implicit curve or portions of a curve. Changing \(C\) changes the level of the relation and generally selects a different member of the family. When an initial point is supplied, substituting that point determines the constant for any member of this family that passes through it.

The relation need not describe a single graph over every possible \(x\)-value. It may have branches, and a branch only counts as a solution on an interval where it is differentiable and where the differential equation is defined. In this example the logarithm also requires \(y\ne0\), while the original equation has a denominator requiring \(y\ne-1\). For verifying a branch of the separated family, both restrictions apply.

Implicit form is useful because it can state a solution family exactly even when isolating \(y\) is inconvenient. To verify the family, it is enough to differentiate its relation and check that the resulting slope agrees with the differential equation wherever the calculation is valid. The relation itself is not a license to divide at a value where the divisor is zero.

A Common Differentiation Mistake

A frequent error is to differentiate \(\ln|y|\) as though \(y\) were the independent variable, forgetting that \(y\) depends on \(x\). The incorrect calculation would be

This loses the chain-rule factor \(dy/dx\). Since \(y=y(x)\), the correct derivative is \(\frac{1}{y}\frac{dy}{dx}\). Thus the correct line is

Factoring and solving then gives \(\frac{dy}{dx}=2xy/(y+1)\), provided \(y\ne0,-1\). A useful check is to differentiate every term involving \(y\) with the chain rule before attempting to isolate the derivative.

Summary

An implicit solution family relates \(x\) and \(y\) without necessarily expressing \(y\) alone. For the family \(\ln|y|+y=x^2+C\), implicit differentiation yields \((1/y+1)\,dy/dx=2x\), and hence the differential equation \(dy/dx=2xy/(y+1)\) where \(y\ne0,-1\). The logarithm restriction \(y\ne0\) belongs to the separated family; it does not remove the solution \(y=0\) from the original differential equation.