Separate variables and retain solutions lost by division
Integrate a separable equation, keep a general constant and check constant solutions before dividing by a dependent-variable factor.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.8. Paper 1 calculus and Paper 3 Integration are assumed. This pathway uses first-order separable equations, including Paper 3 integration methods, with all context assumptions supplied. Second-order equations and integrating-factor methods are beyond this mapped requirement. Check factors before dividing, retain equilibrium solutions, and state the interval and physical meaning of an initial-value solution. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Integrate powers, reciprocals and simple composites.
- Use logarithms on valid domains and substitute into an original derivative equation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is an antiderivative of 1/y on an interval not containing zero?
What you will learn
- Separate a first-order equation into one variable on each side and integrate.
- Identify equilibrium solutions and domain restrictions that division can hide.
Separation turns the equation into two integrals
For dy/dx = f(x)g(y), on an interval where g(y) ≠ 0, write dy/g(y) = f(x) dx and integrate both sides. The resulting relation contains one arbitrary constant; separate constants from both integrals can be combined. Keep an implicit relation when solving explicitly would introduce a branch choice. The procedure is justified along a differentiable solution, not by allowing unrelated values of x and y in each step.
For dy/dx = 2xy, non-zero solutions give dy/y = 2x dx, then ln|y| = x² + C. Absorbing the sign and exponential constant gives y = Ae^(x²), with non-zero A on those branches. The original equation also admits y = 0, omitted when dividing by y. Allowing A = 0 in the final family restores it. Every member has derivative 2xy; different values of A describe different starting conditions.
Check what division and square roots exclude
A zero of g(y) may provide a constant equilibrium solution if the original equation is defined there. Test it directly before dividing. For dy/dx = 3xy², non-zero solutions satisfy −1/y = 3x²/2 + C, while y = 0 is another solution. In contrast, dy/dx = x/y is undefined at y = 0, so zero is not a solution of that equation. Multiplying by y gives y² = x² + C, on branches where y stays non-zero.
After y² = x² + C, an explicit branch is y = ±√(x² + C), valid only where the radicand is positive. The choice of sign cannot change through a point where the original equation is undefined. An equation such as dy/dx = x + y is not directly of the separated product form; writing dy/y = x dx from it would discard the addition. Check algebra before integrating and check the result by differentiation afterwards.
Pause and explain
Which equation permits the constant solution y = 0?
Worked example
Find the general family for dy/dx = 2xy, including any constant solution lost by division.
Show the worked solution
- For y ≠ 0, integrate dy/y = 2x dx to obtain ln|y| = x² + C.
- Write the non-zero branches as y = Ae^(x²), with their sign carried by A.
- Check y = 0 in the original equation; allowing A = 0 includes it in the family.
Answer y = Ae^(x²), A any real constant, including A = 0.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Solve dy/dx = 3xy² in general, retaining any equilibrium solution.
Give me a hint
For non-zero y use ∫y⁻² dy; then check the factor you divided by.
Compare my reasoning
- Separate dy/y² = 3x dx and integrate −1/y = 3x²/2 + C.
- This describes non-zero branches where the denominator of y = −1/(3x²/2 + C) does not vanish.
- Substituting y = 0 directly gives zero on both sides, so it must also be included.
−1/y = 3x²/2 + C for non-zero branches; also y = 0.
Look for these in your work
- Integrate the reciprocal square with its negative sign.
- State both the equilibrium and the non-zero branch restrictions.
Find an implicit general solution of dy/dx = x/y and explain its real branch restrictions.
Give me a hint
Multiply by y only where the original equation is defined.
Compare my reasoning
- Use y dy = x dx, giving y²/2 = x²/2 + C₁.
- Combine constants to obtain y² = x² + C.
- The explicit branches are ±√(x² + C), on intervals with x² + C > 0; y = 0 is excluded by the original equation.
y² = x² + C; choose a non-zero branch where x² + C > 0.
Look for these in your work
- Retain a single arbitrary constant.
- Check the original denominator when stating valid branches.
A supplied model for a non-negative amount y is dy/dt = −2y. Find its general family and explain the permitted constant and whether it can stay zero.
Give me a hint
Separate for positive y, then test zero in the original model.
Compare my reasoning
- For positive y, ln y = −2t + C.
- Thus y = Ae^(−2t); non-negative amounts require A ≥ 0.
- A = 0 gives y = 0 with zero derivative, so a zero initial amount can remain zero under this law.
y = Ae^(−2t), A ≥ 0 for this non-negative model; A = 0 is valid.
Look for these in your work
- Separate the mathematical constant from its physical sign restriction.
- Retain the zero amount when interpreting the family.
Common mistakes
- Check zero factors before dividing by them.
- A denominator in the original equation cannot be set to zero.
- One implicit solution can have different explicit branches on separate intervals.
What can you explain now?
Why must y = 0 be checked before dividing dy/dx = 2xy by y?
Compare with the explanation
Division by y excludes the valid equilibrium y = 0.
The original equation gives derivative zero at that constant function.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Separate a product-form equation, integrate with one constant, and check every divided factor against the original equation before accepting a general solution.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.