Solve separable equations using logarithms partial fractions and inverse tangent
Choose an integration method for each separated side and keep the resulting branch and equilibrium restrictions.
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 positive quadratic reciprocals and linear partial fractions.
- Use principal inverse-tangent values and apply initial conditions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is ∫dx/(4 + x²)?
What you will learn
- Use inverse-tangent and partial-fraction primitives in separated equations.
- Retain equilibria and the interval of an explicit or implicit initial-value solution.
Both separated sides can need Paper 3 methods
For dy/dx = (1 + y²)/(1 + x²), separate dy/(1 + y²) = dx/(1 + x²). Integration gives arctan y = arctan x + C in radians. With y(0) = 1, C = π/4. On the interval containing zero where that angle stays in the principal range, the tangent addition formula gives y = (x + 1)/(1 − x), valid for x < 1. At x = 1 the solution has a pole; the formula at x > 1 belongs to a separate branch and does not continue the initial-value solution through it.
A scaled quadratic changes its primitive: ∫dx/(4 + x²) = (1/2)arctan(x/2), not arctan x. Keep the exact multiplier when it appears in a separated equation. An implicit relation such as y + y³/3 = x² + C can be a complete answer when an elementary explicit formula is unhelpful. Differentiate it to check (1 + y²)y′ = 2x rather than treating y as constant.
Partial fractions also expose excluded equilibria
For dy/dx = y(1 − y), division is valid only away from y = 0 and y = 1. Since 1/[y(1 − y)] = 1/y + 1/(1 − y), integration gives ln|y| − ln|1 − y| = x + C, or ln|y/(1 − y)| = x + C. The second logarithm has a minus sign because the derivative of 1 − y is −1. Retain the original constant solutions 0 and 1 separately.
A non-equilibrium branch can be written y = Ke^x/(1 + Ke^x), where the sign of K is retained and the denominator must stay non-zero. For y(0) = 1/4, K = 1/3 and y = e^x/(3 + e^x), defined for every real x and between 0 and 1. Derivative checking distinguishes the correct logistic expression from an answer with the wrong sign in its denominator. Absolute-value logarithms do not justify dropping a negative multiplicative constant on other branches.
Pause and explain
After dividing dy/dx = y(1 − y) by y(1 − y), which constants must be checked separately?
Worked example
Solve dy/dx = (1 + y²)/(1 + x²) with y(0) = 1 and state the interval containing the initial point.
Show the worked solution
- Integrate the separated reciprocals to get arctan y = arctan x + C.
- The initial value gives C = π/4; apply tangent addition to obtain y = (x + 1)/(1 − x).
- Its derivative is 2/(1 − x)², equal to the original right-hand side; the initial branch has x < 1.
Answer y = (x + 1)/(1 − x), x < 1 for the branch through x = 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 = (1 + y²)/(4 + x²), y(0) = 0.
Give me a hint
Use the scaled inverse-tangent primitive on the x side.
Compare my reasoning
- Integrate dy/(1 + y²) = dx/(4 + x²) to obtain arctan y = (1/2)arctan(x/2) + C.
- At x = 0, y = 0, so C = 0.
- Thus y = tan[(1/2)arctan(x/2)]; the inner angle lies between −π/4 and π/4 for all real x.
y = tan[(1/2)arctan(x/2)], for all real x.
Look for these in your work
- Retain the quadratic scale and its multiplier.
- Check the initial value and that the tangent has no pole on the stated interval.
Solve dy/dx = y(1 − y) with y(0) = 1/4. Also identify the constant solutions of the original equation.
Give me a hint
Use partial fractions and preserve the equilibria excluded by division.
Compare my reasoning
- For non-equilibrium branches, ln|y/(1 − y)| = x + C.
- The initial ratio is 1/3, giving y = e^x/(3 + e^x), which stays strictly between 0 and 1.
- The constant solutions y = 0 and y = 1 also satisfy the original equation but not this initial condition.
y = e^x/(3 + e^x); original equilibria y = 0 and y = 1.
Look for these in your work
- Keep the logarithmic minus sign and initial ratio.
- Distinguish other solutions of the equation from the particular starting condition.
A supplied dimensionless response satisfies dy/dx = 2x/(1 + y²), with y(0) = 1 and x ≥ 0. Find an implicit solution and the first x at which y = 2.
Give me a hint
Integrate (1 + y²) dy = 2x dx and use the starting response.
Compare my reasoning
- Integration gives y + y³/3 = x² + C; the initial value gives C = 4/3.
- At y = 2 the left side is 14/3, so x² = 10/3.
- The first non-negative input is x = √(10/3); the function y + y³/3 is strictly increasing, so it selects one real response at each x.
y + y³/3 = x² + 4/3; reaches y = 2 at x = √(10/3).
Look for these in your work
- Differentiate the implicit relation to recover the rate law.
- Choose the non-negative input and check that the threshold response is unambiguous.
Common mistakes
- Use inverse-trigonometric primitives in radians with their scaling factors.
- The primitive of 1/(1 − y) is −ln|1 − y|.
- Dividing by y(1 − y) hides both equilibrium solutions.
What can you explain now?
What is ∫dy/[y(1 − y)] away from y = 0 and 1?
Compare with the explanation
ln|y| − ln|1 − y| + C.
The partial fractions are 1/y + 1/(1 − y), and the second denominator has derivative −1.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Identify each separated integrand, integrate with the right scaling and signs, and check the particular branch and any equilibria before interpreting the answer.
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.