Why Domain Conditions Matter When Separating
Separating variables is an algebraic rearrangement of a differential equation. The goal is to place expressions involving \(y\) with \(dy\) and expressions involving \(x\) with \(dx\). But rearranging may require dividing by an expression that can equal zero, or working with a logarithm whose argument must be positive.
Those restrictions matter because the rearranged equation may describe only part of the original equation’s solution set. A value excluded by division might still give a valid constant solution. A logarithm may restrict which values are allowed even before any rearrangement takes place.
What Separation Means
A first-order differential equation is separable when it can be written in the form \(y'=g(x)h(y)\), where \(g\) depends only on \(x\) and \(h\) depends only on \(y\). On an interval where \(h(y)\ne 0\), divide by \(h(y)\) and write the separated equation as
This notation places the \(y\)-dependent factor beside \(dy\) and the \(x\)-dependent factor beside \(dx\). The notation is a useful shorthand for a chain-rule relationship; it does not mean that \(dx\) and \(dy\) are ordinary numbers that can be rearranged without conditions.
The condition \(h(y)\ne 0\) is essential for this division. If \(h(c)=0\), then the constant function \(y(x)=c\) satisfies the original equation wherever the equation is defined: its derivative is zero, and \(g(x)h(c)=0\). Such a solution is not included in the division by \(h(y)\), so it must be checked separately.
Why the Rearrangement Works
The chain rule explains the method. Suppose \(y\) is a differentiable solution on an interval and \(h(y(x))\ne 0\) throughout that interval. Define a function \(H\) on the relevant \(y\)-values by requiring \(H'(y)=1/h(y)\). Then
Thus, division by \(h(y)\) is precisely the step that makes the chain rule produce a \(y\)-only factor differentiated through \(y(x)\). The argument applies only where the quotient is defined. It justifies separation on such an interval, not at values where \(h(y)=0\).
In practice, keep three questions in view:
- Where is the original differential equation defined?
- What values are excluded by the division used to separate it?
- Do any excluded values give solutions of the original equation?
Worked Examples: Tracking Restrictions
Worked Example: Dividing by a Factor That Can Be Zero
Separate the equation \(y'=x(y-2)\), keeping track of the value excluded by division.
Step 1: Identify the original domain. The expression \(x(y-2)\) is defined for every real \(x\) and \(y\), so the equation has no domain restriction from its formula.
Step 2: Divide only where permitted. If \(y\ne 2\), divide both sides by \(y-2\):
Using differential notation to place the factors, this is
Step 3: Check the excluded value in the original equation. If \(y(x)=2\), then \(y'=0\) and \(x(y-2)=x(0)=0\). Hence \(y(x)=2\) satisfies the original equation.
Check: On any interval where \(y\ne2\), multiplying the separated equation by \(y-2\) gives \(y'=x(y-2)\), the original equation. The constant solution \(y=2\) is valid but is not represented by the divided equation.
Worked Example: A Rational Factor with Two Kinds of Exclusion
Separate \(y'=\dfrac{xy}{y^2-4}\) and distinguish the equation’s original domain from the restriction added by division.
Step 1: State where the original equation is defined. Its denominator must be nonzero:
Step 2: Separate on values where division by \(y\) is allowed. For \(y\ne0\), multiply the equation by \((y^2-4)/y\):
The separated equation is being used where \(y\ne-2\), \(y\ne2\), and \(y\ne0\). The first two restrictions come from the original equation; the last comes from the division.
Step 3: Check the value excluded only by division. At \(y=0\), the original denominator is \(0^2-4=-4\), which is nonzero. The constant function \(y(x)=0\) has derivative zero, while the right-hand side of the original equation is \(x(0)/(-4)=0\). Therefore \(y=0\) is a solution of the original equation.
Check: Where \(y\ne0,\pm2\), multiplying \(\dfrac{y^2-4}{y}y'=x\) by \(\dfrac{y}{y^2-4}\) recovers \(y'=xy/(y^2-4)\). The equation itself excludes \(y=\pm2\); division additionally excludes \(y=0\), even though that value gives a valid constant solution.
Worked Example: A Logarithm in the Original Equation
Separate \(y'=\dfrac{\ln y}{x}\), and record the restrictions from both the logarithm and the division.
Step 1: Find the original domain. The natural logarithm \(\ln y\) is defined only for \(y>0\), and the denominator requires \(x\ne0\). Thus the original equation is defined only where \(y>0\) and \(x\ne0\).
Step 2: Divide by the \(y\)-dependent factor. To put the \(y\)-factor with \(dy\), divide by \(\ln y\). This requires \(\ln y\ne0\), or \(y\ne1\):
Step 3: Check the excluded value. At \(y=1\), the original right-hand side is \(\ln(1)/x=0\) for \(x\ne0\). The constant function \(y(x)=1\) therefore satisfies the original equation on intervals that do not contain \(x=0\).
Check: For \(y>0\), \(y\ne1\), and \(x\ne0\), multiplying the separated equation by \(\ln y\) recovers \(y'=\ln(y)/x\). The restriction \(y>0\) comes from the original logarithm; \(y\ne1\) is added by division. The constant \(y=1\) solution must be checked in the original equation.
Worked Example: A Denominator That Never Vanishes
Separate \(y'=\dfrac{x+1}{y^2+1}\) and determine whether division introduces an exclusion.
Step 1: Check the denominator. Since \(y^2\ge0\) for real \(y\), \(y^2+1\ge1>0\). The original equation is defined for every real \(x\) and \(y\).
Step 2: Multiply by the denominator. Because \(y^2+1\) is never zero, the rearrangement is valid for every real \(y\):
Check: Dividing the separated equation by \(y^2+1\), which is positive and nonzero, gives \(y'=(x+1)/(y^2+1)\). No real \(y\)-value was lost, and no extra domain restriction was introduced.
Logarithms and Their Domains
Logarithmic expressions need the same care as denominators. For real numbers, \(\ln u\) is defined only when \(u>0\). The expression \(\ln|u|\) is defined only when \(u\ne0\): the absolute value permits either sign of \(u\), but it does not permit \(u=0\).
This distinction is useful when a logarithm appears while manipulating an equation. For example, if \(y\ne0\), then \(\ln(y^2)=2\ln|y|\), because \(y^2>0\) for either sign of \(y\). Replacing this with \(2\ln y\) would be invalid for negative \(y\), since \(\ln y\) is not defined there. Conversely, writing \(\ln|y|\) does not restore the value \(y=0\), where the logarithm remains undefined.
When a differential equation already contains a logarithm, as in the third example, its domain restriction must be stated before separating. If a later algebraic step introduces another logarithm, check that logarithm’s argument separately; do not assume it has the same domain as the original equation.
A Reliable Separation Checklist
Check denominators, logarithms, square roots, and any other expressions that restrict \(x\) or \(y\).
Before dividing by a factor, state that it is nonzero. Keep the separated equation’s restrictions visible.
Substitute a value excluded only by division into the original equation to see whether it gives a constant solution.
On the stated domain, reverse the algebra to confirm that the separated equation is equivalent to the original one.
Common Mistake: Treating the Separated Equation as the Whole Answer
A frequent error is to divide by a factor that can be zero and then forget the value that was excluded. For instance, starting with \(y'=x(y-2)\), a student may write \(\dfrac{1}{y-2}\,dy=x\,dx\) and conclude that \(y=2\) is not a solution because the separated expression is undefined there. That conclusion confuses the domain of the rearranged equation with the domain of the original one.
The correction is to state that the separated equation holds only where \(y\ne2\), then substitute \(y=2\) into the original equation. Since both sides are zero, \(y=2\) is a valid constant solution. The same habit applies to logarithms: replacing \(\ln y\) by \(\ln|y|\) is not a valid way to broaden an equation whose original logarithm requires \(y>0\).
Check Your Understanding
For each question, state the relevant domain restrictions and explain whether division may omit a constant solution.
- Separate \(y'=x(y+3)\). Which value must be excluded during division, and does it satisfy the original equation as a constant solution?
- For \(y'=y/(x-1)\), identify the original restriction on \(x\). What additional restriction appears if you divide by \(y\)?
- For \(y'=\ln(y-2)\), what values of \(y\) are allowed by the original equation? What value makes the right-hand side zero?
- Why is \(\ln|y|\) defined for negative \(y\) but not for \(y=0\)?
- In \(y'=(x+2)/(y^2+4)\), does separating require excluding any real \(y\)-values? Justify your answer.
