What a Separable Differential Equation Asks You to Do
A differential equation describes a relationship between an unknown function and one or more of its derivatives. A separable differential equation is one whose variables can be arranged so that expressions involving the dependent variable are on one side and expressions involving the independent variable are on the other.
For AP free-response questions, solving the equation is only part of the task. A complete response should show the separation, integrate both sides, include the arbitrary constant, and state the solution clearly. The result may be an implicit relation or, when algebra permits, an explicit formula for the dependent variable.
The condition \(g(y)\ne 0\) matters: dividing by \(g(y)\) is not valid when \(g(y)=0\). Any constant solutions arising from those values must be checked separately. Also, the equation itself may be undefined at some values of \(y\), which restricts where a solution can be valid.
A Reliable Solution Format
When solving a separable equation, keep the algebra and the domain conditions visible. A well-organized solution can follow these steps:
Write the derivative as a product of a function of \(x\) and a function of \(y\), if possible.
Find values where the equation is undefined and values where dividing by \(g(y)\) would divide by zero.
Put the \(y\)-expression with \(dy\), the \(x\)-expression with \(dx\), and integrate both sides.
Include an arbitrary constant and give an implicit relation or solve for \(y\) when appropriate. Add any constant solutions excluded by division.
Differentiate the result and confirm that it satisfies the original differential equation on its stated domain.
Writing \(dy\) and \(dx\) in the separated equation is useful bookkeeping. It signals which variable is being integrated on each side; the justification comes from the chain rule, not from treating differentials as ordinary numbers.
Why Separation Works
Suppose \(y(x)\) satisfies \(y'=f(x)g(y)\) on an interval where \(g(y(x))\ne 0\). Choose antiderivatives \(F\) and \(G\) such that \(F'(x)=f(x)\) and \(G'(y)=1/g(y)\). The chain rule gives:
Since \(G(y(x))\) and \(F(x)\) have the same derivative on the interval, they differ by a constant. Therefore \(G(y)=F(x)+C\). This proves why integrating the separated equation produces a solution relation. Conversely, differentiating such a relation recovers the original differential equation wherever the derivatives exist and the divisions used in the derivation are valid.
Worked Example: An Explicit Family of Solutions
Worked Example: An Explicit Family of Solutions
Solve \( \frac{dy}{dx}=\frac{x}{y+1} \), giving an explicit form where possible.
The equation is defined only when \(y\ne -1\). Multiply both sides by \(y+1\), which is nonzero on any interval where a solution is defined:
Write the variables on their respective sides and integrate:
Multiply by \(2\), absorbing the factor of \(2\) into the arbitrary constant, and complete the square:
Thus an implicit general solution is \( (y+1)^2=x^2+C_2 \). Where \(x^2+C_2>0\), this can be written explicitly as
To check, differentiate the implicit relation on an interval where \(y\ne -1\):
This is the original equation. The explicit branches also satisfy it wherever the square root is positive and the derivative exists. The excluded value \(y=-1\) is not a solution because the original equation is undefined there.
Implicit Solutions Are Complete Answers
After integration, it is not always possible—or useful—to isolate \(y\). An equation relating \(x\) and \(y\) can be a valid implicit general solution if it accurately describes the solution family. If you do isolate \(y\), check whether algebra introduces separate branches and whether the resulting formulas remain in the domain of the original equation.
Worked Example: An Implicit Solution
Solve \( \frac{dy}{dx}=\frac{x}{1+y^2} \) and verify the implicit result.
The denominator \(1+y^2\) is positive for every real \(y\), so separation does not exclude any real \(y\)-values:
Integrate each side term by term:
This implicit relation is the general solution. To verify it, differentiate both sides with respect to \(x\), using the chain rule on the left:
The final equation is the original differential equation. No division by a factor that can equal zero was required, so this separation has not omitted an equilibrium solution.
Check for Solutions Lost by Division
When the equation contains a factor depending on \(y\), dividing by that factor can remove a valid constant solution. A function \(y(x)=k\) is constant when \(y'=0\). Substituting it into \(y'=f(x)g(y)\) shows that it is a solution if \(f(x)g(k)=0\) for every \(x\) in the interval. In particular, if \(g(k)=0\), the constant function \(y=k\) satisfies the equation.
Worked Example: Finding Equilibrium Solutions
Solve \( \frac{dy}{dx}=(x+1)(y^2-4) \), including any constant solutions omitted during separation.
First find values that make the \(y\)-factor zero:
Substitution into the original equation shows that \(y(x)=2\) and \(y(x)=-2\) are both constant solutions: in each case, \(y'=0\) and the right-hand side is also \(0\).
For nonconstant solutions with \(y\ne 2\) and \(y\ne -2\), separate variables:
Factor the denominator and use partial fractions:
Integrating gives:
To check this implicit solution, differentiate both sides:
The difference in parentheses simplifies to \(4/(y^2-4)\). Therefore
as required. This logarithmic family applies where \(y\ne \pm2\); the two constant solutions found before division must be listed separately.
A Common Pitfall: Reporting Only the Divided Equation
A tempting but incomplete response to the previous example is to divide by \(y^2-4\), integrate, and report only the logarithmic relation. The division is valid only for \(y\ne 2\) and \(y\ne -2\), so that relation does not include the constant solutions.
The correction is to test those values in the original equation before or immediately after dividing. At \(y=2\), the right-hand side is \((x+1)(4-4)=0\), matching the derivative of the constant function. At \(y=-2\), it is \((x+1)(4-4)=0\) as well. Both constants therefore belong in the answer.
Presenting the Answer in an FRQ
A free-response solution should make the mathematical logic easy to follow. Show the separated equation before integrating, keep the arbitrary constant in the integrated relation, and state whether the answer is implicit or explicit. If you divided by an expression involving \(y\), identify the values excluded by that division and check whether they give constant solutions.
Finally, verify the result by differentiating. This check is especially useful for implicit answers: apply the chain rule to the \(y\)-terms, then rearrange until the original differential equation appears. State any domain restrictions needed for that differentiation or for the original equation to be defined.
Check Your Understanding
For each question, show enough work to make the separation and the role of any excluded values clear.
- Separate and integrate \( \frac{dy}{dx}=x(1+y^2) \). Give an implicit general solution.
- For \( \frac{dy}{dx}=y(y-3) \), which constant solutions might be omitted if you divide by the \(y\)-dependent factor?
- Why is an implicit relation an acceptable solution when it cannot easily be solved for \(y\)?
- What condition must hold before dividing both sides by \(g(y)\) in an equation \(y'=f(x)g(y)\)?
- How can differentiating an implicit solution check that it satisfies the original differential equation?
