Why Separation Is a Derivation, Not a Trick
A first-order differential equation describes how a function changes. For an equation whose right-hand side factors into an expression in \(x\) and an expression in \(y\), it is natural to try to arrange the \(y\)-dependence on one side and the \(x\)-dependence on the other. This arrangement is called separation of variables.
The rearrangement is useful, but it has a condition: dividing by the factor that depends on \(y\) is valid only where that factor is nonzero. The condition matters because the original equation may still make sense at a value where the factor is zero, even though the divided expression does not.
This tutorial derives the rearrangement and explains its restriction. It does not yet carry out the later step of integrating the separated sides.
The Algebra Behind the Separated Form
Let \(y=y(x)\) be a differentiable function on an interval \(I\), and suppose it satisfies
Here \(f\) is defined for the \(x\)-values under consideration, and \(g\) is defined at the values taken by \(y\). Suppose, in addition, that \(g(y(x))\ne 0\) for every \(x\in I\). Then division by \(g(y(x))\) is valid at each point of the interval:
Writing \(dy/dx\) for the derivative, the expression on the left is commonly recorded in differential notation as \(dy/g(y)\), while the right is recorded as \(f(x)\,dx\). Thus the separated form is
This notation is shorthand for a statement about derivatives. It does not require treating \(dy\) and \(dx\) as ordinary numbers that can be independently cancelled. Along the function \(y(x)\), the precise derivative statement represented by the separated form is
In particular, the condition \(g(y(x))\ne 0\) is exactly what permits the displayed quotient to be defined. The separated form is not an instruction to divide at values where its denominator vanishes.
Why the Rearrangement Preserves the Differential Equation
The algebraic justification is reversible wherever the denominator is nonzero. Starting from the original equation, division by \(g(y(x))\) gives the separated derivative statement. Conversely, multiplying that statement by \(g(y(x))\) recovers the original equation. So, on an interval where \(g(y(x))\ne 0\), the two statements describe the same derivative condition.
Proof. If the original equation holds, divide both sides by \(g(y(x))\), which is allowed because it is nonzero. This gives \(y'(x)/g(y(x))=f(x)\). In the reverse direction, multiply both sides of that equality by \(g(y(x))\). The result is \(y'(x)=f(x)g(y(x))\). Both steps are valid at every \(x\in I\) by the stated nonzero condition. This proves the equivalence on \(I\).
The chain rule gives a second way to understand why the separated form is useful. Suppose \(g\) is continuous and nonzero on an interval \(J\) containing the values of \(y\). On \(J\), define an antiderivative \(G\) of \(1/g\), so \(G'(u)=1/g(u)\). If \(y'=fg(y)\), then
The cancellation in this chain-rule calculation is valid because \(g(y(x))\ne0\). It shows that separation converts the original equation into a derivative involving \(x\) alone. Finding antiderivatives and using this relation to obtain solution formulas are further steps; the derivation here explains why the rearrangement is justified.
Worked Examples of the Rearrangement
Worked Example: A Linear Factor in \(y\)
Consider the equation
Here \(f(x)=2x+1\) and \(g(y)=y+3\). The division step requires \(y+3\ne0\), or \(y\ne-3\). On any interval where a solution avoids \(-3\), divide the equation by \(y+3\):
In differential notation this is
Check the rearrangement at \(x=2\) and \(y=1\). The original right-hand side is \((2\cdot2+1)(1+3)=5\cdot4=20\). The original equation therefore requires \(dy/dx=20\) at that point. The left side of the separated derivative statement is \(20/(1+3)=20/4=5\), while its right side is \(2\cdot2+1=5\). The values agree, so the separated statement is consistent with the original equation at this point. The check uses \(y=1\), for which the denominator \(y+3=4\) is nonzero.
Worked Example: A Trigonometric Factor
Consider
Here \(f(x)=x^2\) and \(g(y)=\sin y\). Division by \(g(y)\) requires \(\sin y\ne0\). Since \(\sin y=0\) when \(y=k\pi\) for an integer \(k\), the separation step applies on intervals where \(y\) avoids those values. Dividing where this condition holds gives
or, in differential notation,
Check at \(x=1\) and \(y=\pi/2\). The original right-hand side is \(1^2\sin(\pi/2)=1\cdot1=1\), so the derivative must be \(1\). In the separated derivative statement, the left side is \(1/\sin(\pi/2)=1/1=1\), and the right side is \(1^2=1\). The equality checks exactly, and the division is permitted because \(\sin(\pi/2)=1\ne0\).
Worked Example: A Quadratic Factor with Two Excluded Values
Consider
The factor depending on \(y\) is \(g(y)=y^2-4=(y-2)(y+2)\). It vanishes at \(y=2\) and \(y=-2\). Therefore, division by this factor is valid only where \(y\ne2\) and \(y\ne-2\). On an interval where both conditions hold, the rearrangement is
or
For a numerical check, take \(x=3\) and \(y=0\). The original right-hand side is \((3-2)(0^2-4)=1\cdot(-4)=-4\), so \(dy/dx=-4\) there. The left side of the separated derivative statement is \((-4)/(0^2-4)=(-4)/(-4)=1\), and the right side is \(3-2=1\). The two sides match. Also, \(y=0\) is allowed in this division because \(0\) is neither excluded value.
Where the Restriction Comes From
The restriction is not an extra rule added for convenience. It follows directly from the arithmetic of the derivation: division by zero is undefined. If \(g(y(x))=0\) at a point, then the quotient \(y'(x)/g(y(x))\) is not defined there, so the separated derivative statement cannot be obtained at that point by division.
This does not mean that the original differential equation is undefined there. In the original equation, the factor \(g(y(x))\) is multiplied by \(f(x)\), and that product may be perfectly well-defined when \(g(y(x))=0\). The correct conclusion is limited: the division step does not apply at such a point. Any analysis of the original equation at those values must not rely on the divided equation.
When a solution remains in a region where \(g\ne0\), the separation is valid throughout that interval. If a solution reaches a value where \(g=0\), the quotient form is unavailable at that point; one cannot claim that the division established an equivalent equation there. This interval-based statement is the precise scope of the method being derived.
Common Pitfall: Dividing at a Zero
For example, start with
A careless rearrangement might divide by \(y+3\) and then claim the result applies for every value of \(y\):
This claim is wrong. At \(y=-3\), its left-hand side has denominator \(-3+3=0\), so it is undefined. The original equation, by contrast, has right-hand side \((x+1)(-3+3)=(x+1)\cdot0=0\), which is defined. The correction is to state the condition: the divided form is valid only where \(y+3\ne0\), that is, where \(y\ne-3\). The original equation must be treated separately at values excluded by the division.
Check Your Understanding
For each question, focus on the validity and meaning of the separation step, not on finding a solution formula.
- For \(dy/dx=(x^2+1)(y-5)\), what condition on \(y\) permits division, and what is the separated differential form?
- For \(dy/dx=e^x\cos y\), at which \(y\)-values is division by the \(y\)-factor not permitted?
- Explain why \(dy/g(y)=f(x)\,dx\) is shorthand for a derivative statement rather than a license to treat \(dy\) and \(dx\) as ordinary numbers.
- Use the chain rule to show why \(dG(y(x))/dx=f(x)\) when \(G'(y)=1/g(y)\) and \(y'=f(x)g(y)\), assuming \(g(y(x))\ne0\).
- In the equation \(dy/dx=x(y^2-9)\), identify the \(y\)-values excluded by division and state why the original equation can still be defined at those values.
