Why Careful Separation Matters
A separable differential equation often becomes easier to analyze after its \(x\)- and \(y\)-dependence are placed on opposite sides. But this algebra has a condition: whenever you divide by an expression, that expression must be nonzero at the points where the division is used.
This matters because a factor can vanish for an entire solution, or at particular points in its domain. Dividing by that factor can exclude a solution or make an equation undefined where the original equation was meaningful. A correct separation therefore records not only the resulting equation, but also the conditions under which the algebraic steps preserve equivalence.
What Equivalence Means Here
Consider a differential equation on an interval \(I\), with an unknown differentiable function \(y:I\to\mathbb{R}\). Two equations are equivalent on \(I\) if they have exactly the same differentiable solutions on that interval, subject to the same stated restrictions.
For a product-form equation \(y'=g(x)h(y)\), one might divide by \(h(y)\) to obtain \(y'/h(y)=g(x)\). This rearrangement is equivalent at a point \(x\) only if \(h(y(x))\ne 0\). Thus, on an interval where \(h(y(x))\) never vanishes, the two equations have the same solutions. If \(h(y(x))=0\) somewhere, the divided equation is not defined there, even though the original equation may be.
A Safe Separation for the Focus Equation
Worked Example: Separating Without Dividing by a Vanishing Factor
Consider \(y'=(x+1)(y^2+1)\). The dependent-variable factor satisfies \(y^2+1\ge 1>0\) for every real \(y\). Division by this factor is therefore valid at every point of any real-valued solution.
Divide both sides by \(y^2+1\):
The cancellation is valid because \(y^2+1\ne 0\). Reversing the step confirms the equivalence: multiply both sides by \(y^2+1\) to recover \(y'=(x+1)(y^2+1)\).
At the point \(x=-1,\ y=2\), the original right-hand side is \((-1+1)(2^2+1)=0\cdot5=0\). The separated right-hand side is also \(-1+1=0\), and the separated left-hand side is \(y'/5\), which equals \(0\) when the original equation gives \(y'=0\). The separated equation remains defined at \(x=-1\).
Notice that no division by \(x+1\) was needed. That factor vanishes at \(x=-1\), so dividing by it would make the rearranged equation undefined there. Keeping it on the right-hand side avoids that unnecessary restriction.
Division Can Exclude a Solution
If \(h(c)=0\) for some constant \(c\), the constant function \(y(x)=c\) satisfies \(y'=g(x)h(y)\): its derivative is zero, and the right-hand side is \(g(x)h(c)=0\). Dividing by \(h(y)\) is not valid for this solution. It must be considered separately rather than assumed to be included in the divided equation.
Worked Example: A Zero Factor and a Constant Solution
Consider \(y'=xy\). Dividing by \(y\) gives \(y'/y=x\), but this division requires \(y\ne 0\). First check the excluded case: if \(y(x)=0\), then \(y'=0\) and \(xy=x\cdot0=0\), so \(y=0\) is a solution of the original equation.
For a solution branch on an interval where \(y\ne0\), division is valid:
To check the algebra at \(x=2,\ y=3\), the original right-hand side is \(2\cdot3=6\). If \(y'=6\), the separated left-hand side is \(6/3=2\), which equals \(x\). Multiplying the separated equation by \(y=3\) recovers \(y'=xy=6\). The constant solution \(y=0\) is not represented by this divided equation, because \(y'/y\) is undefined there.
Division does not prove that every other solution must stay away from zero on its entire domain. It says only that the displayed divided equation is valid on intervals where \(y\ne0\). Any claim about whether a solution can reach or cross a zero requires further reasoning.
Do Not Divide by a Factor You Do Not Need
Worked Example: Keeping the Independent-Variable Factor
Consider \(y'=(x-2)(y-3)\). The factor \(y-3\) can be divided out only on an interval where \(y\ne3\). The factor \(x-2\), however, need not be divided out at all.
For a branch with \(y\ne3\), divide only by \(y-3\):
Reverse the step by multiplying by \(y-3\), which gives the original equation. At \(x=4,\ y=5\), the original right-hand side is \((4-2)(5-3)=2\cdot2=4\). If \(y'=4\), the separated left-hand side is \(4/(5-3)=4/2=2\), equal to \(x-2=4-2=2\).
The excluded constant case is \(y=3\). Substitution into the original equation gives \(y'=0\) and \((x-2)(3-3)=0\), so \(y=3\) is indeed a solution. Dividing by \(x-2\) as well would be unnecessary and would make that rearrangement undefined at \(x=2\), even though the original equation is defined there.
Why the Chain Rule Supports Separation
Separation is often followed by finding antiderivatives. The relevant chain-rule justification needs a regularity condition: nonvanishing alone does not guarantee that a function has an antiderivative. Suppose \(g\) and \(h\) are continuous, \(J\) is an interval on which \(h(y)\ne0\), and \(y(x)\in J\) throughout an interval \(I\). Then \(1/h\) is continuous on \(J\), so it has an antiderivative \(H\) there. By definition, \(H'(u)=1/h(u)\) for \(u\in J\).
Proof. Suppose first that \(y'=g(x)h(y)\). The chain rule gives
The cancellation is permitted because \(h(y(x))\ne0\) on \(I\). Conversely, suppose \(\dfrac{d}{dx}H(y(x))=g(x)\). Applying the chain rule and using \(H'=1/h\) gives \(y'(x)/h(y(x))=g(x)\). Multiplication by the nonzero quantity \(h(y(x))\) yields \(y'(x)=g(x)h(y(x))\). This proves both directions on the stated interval. If \(h\) has zeros, this argument applies only on intervals where the solution’s values avoid them; solutions at the zeros must be checked in the original equation.
A Common Mistake
A tempting but invalid step for \(y'=(x+1)(y^2+1)\) is to divide both sides by \(x+1\) and write \(y'/(x+1)=y^2+1\) as an equivalent equation for every real \(x\). At \(x=-1\), the quotient \(y'/ (x+1)\) is undefined. The original equation, by contrast, is defined there and says \(y'(-1)=0\). The divided equation therefore cannot be asserted on an interval containing \(-1\).
A Reliable Check Before Separating
Write the equation in product form, such as \(y'=g(x)h(y)\), and note which factors could be zero.
State where it is nonzero. If it can vanish, record the excluded case and test it in the original equation.
Do not claim an equation is valid at points where a denominator is zero. Avoid dividing by factors that do not need to be removed.
Multiply by the stated nonzero divisor to verify that the original equation follows on the same interval.
Why This Careful Bookkeeping Matters
A separated equation is useful only if its connection to the original equation is understood. Dividing by a factor that vanishes may remove a constant solution, restrict the interval under consideration, or leave an equation undefined at points where the original equation still makes sense. Stating the conditions prevents an algebraic rearrangement from being mistaken for a fully equivalent equation.
The key habit is simple: before dividing, ask where the divisor is zero. If it never vanishes, the step is safe throughout the stated domain. If it can vanish, separate the nonzero case from the excluded case and check the latter in the original differential equation. Then reverse the algebra to confirm the claimed equivalence.
Check Your Understanding
For each question, state the conditions that make the algebra valid.
- In \(y'=(x+1)(y^2+1)\), why can you divide by \(y^2+1\) for every real-valued solution?
- For \(y'=xy\), what solution must be checked separately before dividing by \(y\)?
- Why does dividing by \(x-2\) in \(y'=(x-2)(y-3)\) restrict where the rearranged equation is defined?
- What continuity assumptions ensure that \(1/h\) has an antiderivative on an interval where \(h\) is nonzero?
- What operation should you perform to check that a separated equation implies the original equation?
