Why Division Needs Care
Separation of variables is useful because it can turn a differential equation into an equation that can be integrated. But a division used to separate variables may also change which solutions are being considered. In particular, dividing by a factor involving \(y\) is not valid where that factor is zero.
Those zero values deserve attention before any division. They may give constant solutions, also called equilibrium solutions. If they are not checked in the original equation first, they can disappear from the solution work.
For example, the equation \(y'=y(y-1)\) has right-hand side zero when \(y=0\) or \(y=1\). Dividing by \(y(y-1)\) assumes that \(y\) is neither of those values. The two constant solutions therefore need to be handled separately from the divided equation.
Equilibrium Solutions
A solution is a differentiable function \(y(x)\) that satisfies a differential equation on an interval. A constant solution has the form \(y(x)=c\) for one fixed number \(c\). Its derivative is \(y'(x)=0\) everywhere on the interval.
The phrase “for every \(x\)” matters for a nonautonomous equation: the right-hand side must vanish along the entire constant function, not merely at one point. For example, if \(f(x,y)=x(y-3)\), then \(y=3\) is an equilibrium because \(f(x,3)=0\) for every \(x\).
For an autonomous equation, finding equilibria amounts to finding the values \(c\) for which the right-hand side is zero. If the right-hand side is factored, its factors make these values easy to inspect. This step is not merely an optional shortcut: it identifies values at which a later division could fail.
What Division Does to an Equation
Suppose the original equation can be written as
At any point where \(H(y)\ne 0\), dividing both sides by \(H(y)\) gives
That algebraic step is valid only where \(H(y)\) is nonzero. It cannot establish what happens on a solution for which \(H(y)=0\). If a value \(c\) satisfies \(H(c)=0\), it is a candidate equilibrium: substitute \(y=c\) into the original right-hand side and check whether it vanishes for every \(x\). If it does, \(y(x)=c\) is a solution.
Conversely, zeros of a factor are candidates, not automatic equilibria in every possible factorization. The original equation determines whether a constant works. This distinction is especially important if other factors or expressions are not defined at a candidate value.
A Derivation of the Safe Separation Step
The reason for checking the nonzero condition is visible in the chain rule. On an interval where \(H(y(x))\ne 0\), define an antiderivative \(F\) by \(F'(y)=1/H(y)\). Then
This justifies integrating the separated equation on that interval. But \(F'(y)=1/H(y)\) is not defined at a zero of \(H\), and the division in the calculation is not valid there. Thus this derivation proves the separated relation only on intervals where \(H(y(x))\ne0\). It does not rule out solutions that remain at a zero of \(H\).
The correct workflow follows directly from that restriction: first find and test constant solutions in the original equation; then divide only on the remaining parts of the solution work, stating that the relevant factor is nonzero there.
Worked Example: The Equation \(y'=y(y-1)\)
Worked Example: The Equation \(y'=y(y-1)\)
Find the equilibrium solutions, then separate variables for the nonconstant cases.
Step 1: Find the constant solutions before dividing. For \(y(x)=c\), the derivative is zero. The original equation requires
The product is zero when \(c=0\) or \(c=1\). Direct substitution confirms both: \(y=0\) gives \(y'=0=0(0-1)\), and \(y=1\) gives \(y'=0=1(1-1)\). Thus \(y=0\) and \(y=1\) are equilibrium solutions.
Step 2: Separate only where \(y\ne0\) and \(y\ne1\). On any interval where these conditions hold,
The partial fraction identity is
since \(-1/y+1/(y-1)=[-(y-1)+y]/[y(y-1)]=1/[y(y-1)]\). Integrating both sides gives
or
Exponentiating and absorbing the sign into a constant \(K\) gives \((y-1)/y=Ke^x\). Because this equation came from logarithms of nonzero quantities, this separated work applies to the nonconstant branch with \(K\ne0\). Solving for \(y\),
This formula is considered on intervals where its denominator is nonzero. To check it, let \(D=1-Ke^x\). Then \(D'=-Ke^x\), so
Also \(y-1=Ke^x/(1-Ke^x)\), and therefore
The nonconstant formula satisfies the original equation. The separately checked constants \(0\) and \(1\) do as well. Neither should be discarded just because they were excluded during division and logarithmic integration.
Worked Example: A Repeated Equilibrium Factor
Worked Example: A Repeated Equilibrium Factor
Consider \(y'=(y-4)^2\). Find the equilibrium and the separated nonconstant solutions.
Step 1: Test constant functions in the original equation. For \(y=c\), the equation requires \(0=(c-4)^2\), so \(c=4\). Substitution gives \(y'=0=(4-4)^2\), confirming that \(y=4\) is an equilibrium.
Step 2: Divide only when \(y\ne4\). For a nonconstant branch on an interval where \(y-4\ne0\),
Let \(u=y-4\), so \(du=dy\). Integrating gives
Returning to \(y\), \(-1/(y-4)=x+C\). Rearranging,
This branch is valid where \(x+C\ne0\). Its derivative is \(y'=1/(x+C)^2\), while \(y-4=-1/(x+C)\), so \((y-4)^2=1/(x+C)^2=y'\). The constant solution \(y=4\), lost at the division step, also satisfies the original equation.
Worked Example: A Nonautonomous Equation
Worked Example: A Nonautonomous Equation
Consider \(y'=(x+1)(y-3)\). Find the constant solution and solve by separation where division is allowed.
Step 1: Test a constant \(y=c\). The equation requires \(0=(x+1)(c-3)\) for every \(x\). Since \(x+1\) is not zero for every \(x\), this identity holds for all \(x\) only if \(c=3\). Substitution verifies \(y=3\): its derivative is zero, and the right-hand side is \((x+1)(3-3)=0\).
Step 2: Separate when \(y\ne3\). On an interval where the factor \(y-3\) does not vanish,
Integrating,
Exponentiating and absorbing the sign into a constant \(A\) gives
The separated derivation describes the nonconstant branch, for which \(A\ne0\). To check it, differentiate:
The right-hand side of the original equation is
Thus the nonconstant formula satisfies the equation, and the separately tested constant \(y=3\) does too.
A Common Mistake
A tempting but incorrect conclusion for \(y'=y(y-1)\) is: “Divide by \(y(y-1)\), solve the separated equation, and the resulting formula gives every solution.” The division was performed under the conditions \(y\ne0\) and \(y\ne1\). It cannot account for a solution that is constantly zero or constantly one. For instance, after division the equation \(y'=y(y-1)\) becomes \(y'/[y(y-1)]=1\); substituting \(y=0\) into this divided equation would require division by zero.
The correction is not to force the excluded values into the separated calculation. Check them in the original equation first, record the equilibria \(y=0\) and \(y=1\), and then solve the divided equation only on the cases where its denominator is nonzero. Finally, verify each resulting branch in the original equation.
A Reliable Order of Work
Identify values of \(y\) that make the right-hand side zero for every \(x\), and test the corresponding constant functions.
If dividing by \(H(y)\), specify that the separated calculation applies only where \(H(y)\ne0\).
Keep the nonzero conditions in mind while using algebra, logarithms, or other antiderivatives.
Differentiate each proposed nonconstant solution and compare with the original right-hand side; substitute each equilibrium directly as well.
Why This Distinction Matters
A separated formula describes solutions only under the conditions used to derive it. Dividing by a factor is not a harmless rearrangement at points where that factor vanishes. Recording equilibrium solutions first keeps the answer faithful to the original differential equation and makes the limitations of the nonconstant calculation explicit.
This is also why checking by differentiation is valuable: a proposed formula can be tested directly in the equation, while a constant solution can be checked by substituting its value. A complete solution account distinguishes the branches found by separation from the equilibrium solutions that division excluded.
Check Your Understanding
For each question, pay attention to which equation is being tested: the original equation or one obtained after division.
- For \(y'=y(y-2)\), which constant functions should be tested before dividing by \(y(y-2)\)?
- In \(y'=(x+2)(y+1)\), what condition must a constant \(c\) satisfy for \(y=c\) to be an equilibrium for every \(x\)?
- Why is \(\frac{y'}{y-4}=x\) not equivalent to \(y'=x(y-4)\) at values where \(y=4\)?
- When \(y'=y^2\), what restriction must be stated before separating by dividing by \(y^2\)?
- After finding a nonconstant formula by separation, what two checks help ensure that the solution account is complete?
