Why Rearrangement Requires Care
A differential equation may not look separable at first. Its numerator and denominator can hide factors that, once exposed, make the dependence on \(x\) and \(y\) easier to distinguish. But algebraic simplification is not automatically harmless: an equation is defined only where its original denominator is nonzero, and canceling a factor can change that domain.
For a differential equation written as \(y'=F(x,y)\), a solution on an interval is a differentiable function \(y=y(x)\) whose graph lies in the domain of \(F\) and for which \(y'(x)=F(x,y(x))\) at every point of that interval. Thus, rearranging the formula must preserve not only its numerical values where the original expression is defined, but also the conditions under which it is defined.
A Careful Approach to Algebraic Rearrangement
Suppose an equation has a rational right-hand side involving both variables. Factoring may reveal a product of a function of \(x\) and a function of \(y\), or expose factors that can be canceled. Once the expression is in a suitable form, separation means collecting the \(y\)-dependent factors with \(dy\) and the \(x\)-dependent factors with \(dx\). This step can require dividing by a factor that might be zero, so any solutions on its zero set must be considered separately.
There are two different domain issues to track:
- Original denominator restrictions: values where the original right-hand side is undefined are not part of the domain of the differential equation.
- Division restrictions: dividing by a factor requires that factor to be nonzero. If it can be zero along a solution, that case needs separate attention.
Cancellation introduces a related issue. If a factor occurs in both numerator and denominator, canceling it is valid only away from its zeros. The simplified formula may be defined at additional points, but those points do not automatically become allowable for the original equation.
Worked Example: Cancel a Common Factor Without Forgetting the Domain
Worked Example: Cancel a Common Factor Without Forgetting the Domain
Consider the differential equation
The original denominator requires \(y\ne -1\). Factor the numerator:
Therefore, on the original domain \(y\ne -1\), cancellation gives
Now separate and integrate. Since the right-hand side depends only on \(x\), write \(dy=x\,dx\). Integrating both sides gives
To check, differentiate the result: \(y'=x\). Substitution into the original equation, at points where \(y\ne -1\), gives
This verifies the equation wherever its original denominator is nonzero. The reduced equation \(y'=x\) also allows a curve to pass through \(y=-1\), but the original equation is undefined at such a point. Consequently, only portions of the formula \(y=x^2/2+C\) that stay in the original domain can be solutions of the original equation on their entire interval.
What Cancellation Does to an Equation
The previous example illustrates a general fact. Suppose a right-hand side contains a factor \(G(x,y)\) in both its numerator and denominator. Where \(G(x,y)\ne 0\), the original and canceled expressions agree. At points where \(G(x,y)=0\), however, the original expression may be undefined even though the canceled expression has a value.
Cancellation and division also have different effects to watch for. Canceling a factor from numerator and denominator can make a formerly forbidden point appear allowed. Dividing both sides by a factor can discard cases in which that factor is zero. Neither operation justifies silently changing the set of solutions.
Worked Example: Cancellation Leaves Another Restriction
Worked Example: Cancellation Leaves Another Restriction
Consider
The original denominator is zero when \(y=1\) or \(y=-1\), so both values are excluded. Cancel the factor \(y+1\), but retain the restrictions inherited from the original equation:
Multiply by \(y-1\) to separate the variables:
Using \(y'\,dx=dy\), integrate both sides:
For a check, differentiate the implicit relation with respect to \(x\):
Since \(y\ne1\), division by \(y-1\) gives \(y'=x^2/(y-1)\), which is the canceled equation. At points also satisfying \(y\ne-1\), this equals the original right-hand side because
The check confirms the differential relation where the original equation is defined. The value \(y=-1\) remains excluded even though the canceled expression alone is defined there. A solution of the original equation must stay in its original domain throughout its interval.
When Division Can Lose a Solution
To separate variables, one often divides by a factor involving \(y\). Before doing so, ask whether that factor could be zero along a solution. If it can, test that case directly in the original equation. A constant solution, when valid, is called an equilibrium solution; the earlier material on equilibrium solutions lost by division explains this issue in more depth.
Here is the algebraic point in a simple form. If an equation leads to \(A(y)y'=B(x)C(y)\), dividing by \(C(y)\) assumes \(C(y)\ne0\). Any case with \(C(y)=0\) is omitted by that division. It may fail to satisfy the original equation, or it may give a solution that must be recorded separately. The original equation decides which is true.
Worked Example: Factor Before Separating
Worked Example: Factor Before Separating
Consider the equation
Factor the numerator and denominator:
The original denominator imposes \(y\ne0\) and \(x\ne-1\). The equation becomes
First check the case \(y=-1\), because a factor \(y+1\) appears in the numerator. For the constant function \(y(x)=-1\), the derivative is zero, and, when \(x\ne-1\), the original right-hand side is
Thus \(y=-1\) is a solution on intervals that do not contain \(x=-1\). It must not be lost by dividing by \(y+1\).
For solutions with \(y\ne-1\), multiply by \(y/(y+1)\) to separate:
Since \(\frac{y}{y+1}=1-\frac{1}{y+1}\), integrate each side:
To check the implicit relation, differentiate both sides with respect to \(x\):
so
For \(y\ne0,-1\) and \(x\ne-1\), solving this relation for \(y'\) gives \(y'=x(y+1)/(y(x+1))\), the original equation. The logarithmic formula applies on intervals where its logarithms are defined and the original domain restrictions hold. Together with the separately checked solution \(y=-1\), this accounts for the division case in this example.
Why the Rearrangement Argument Works
The key justification is local: every algebraic step is an equivalence only under its stated nonzero conditions. On a region where the canceled factors and denominators are nonzero, the original expression and the rearranged expression have the same value. Multiplying or dividing by a nonzero factor preserves equality there. Integration then produces a relation whose derivative recovers the separated equation, and the domain conditions allow that equation to be translated back to the original one.
Identify every value that makes the original denominator zero before simplifying.
Cancel common factors only on the set where those factors are nonzero, while retaining the original exclusions.
Test its zero case in the original equation rather than assuming it cannot occur.
Differentiate the resulting relation and check that it gives the original equation on the original domain.
This is why algebraic rearrangement is more than cosmetic. A simplified equation can be easier to solve, but its domain may differ from the original equation’s domain. Keeping the conditions attached to each step lets you use the simpler form without claiming solutions at points where the original equation is not defined.
A Common Mistake
A frequent error is to cancel a factor and then treat the simplified expression as if it had always been the original equation. For example, one might write
for every \(y\). That statement is false as an equality of the original expressions at \(y=-1\): the left-hand side is undefined there. The correction is to write
Another error is to divide by \(y+1\) without checking \(y=-1\). In the third example, doing so would remove the valid constant solution \(y=-1\). The remedy is to check the zero case in the original equation before proceeding with the divided equation.
Check Your Understanding
For each question, pay attention to both the algebra and the domain on which the original equation is defined.
- For \(y'=\frac{x(y+2)}{y+2}\), what restriction comes from the original denominator, and what equation results after cancellation on its valid domain?
- Why must \(y=-1\) remain excluded in the original equation \(y'=\frac{x^2(y+1)}{(y-1)(y+1)}\), even after the factor \(y+1\) is canceled?
- In the equation \(y'=\frac{x(y+1)}{y(x+1)}\), why should \(y=-1\) be checked before dividing by \(y+1\)?
- What does differentiating an implicit solution relation check, and why must the original domain restrictions still be applied afterward?
- State the condition under which canceling a common factor preserves equivalence between the original and simplified expressions.
