Apply an initial condition and verify the solution branch
Use a stated starting point to determine a constant, choose the correct sign branch and verify the derivative and initial value.
Section 3.8 · Differential equations
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
- Separate variables and integrate with one constant.
- Differentiate an implicit relation or a composite square root.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If y² = x² + C and y(0) = 2, what is C?
What you will learn
- Use an initial condition to select a particular solution.
- Check the chosen branch in both the original equation and the starting condition.
An initial condition determines a member of the family
For dy/dx = x/y, integration gives y² = x² + C. If y(0) = 2, substitution gives C = 4. The positive initial value selects y = √(x² + 4); the negative square root has the same implicit relation but fails the starting value. Differentiating gives y′ = x/√(x² + 4) = x/y, verifying the original equation. The radicand is positive for every real x, so this branch has no zero-denominator boundary.
A stated initial condition may occur at any x or t, not only zero. For dy/dx = 2x/y with y(1) = −3, integration gives y² = 2x² + C. Substitution gives C = 7 and the negative branch y = −√(2x² + 7). Its derivative is −2x/√(2x² + 7), which equals 2x/y. Squaring the initial value finds C but does not retain its sign; use the original condition again when choosing the branch.
Verify the function on its interval, not at one point only
For dy/dt = −y² with y(0) = 2, non-zero separation gives 1/y = t + C and C = 1/2. Thus y = 2/(1 + 2t). Its derivative −4/(1 + 2t)² equals −y² wherever the denominator is non-zero. The interval containing the initial point is t > −1/2; a forward elapsed-time model uses t ≥ 0. A formula on the other side of a pole does not extend this initial-value solution through the undefined point.
Use three checks: substitute the starting condition, differentiate the proposed function and compare with the original right-hand side, and state an interval where both are defined. If a solution is implicit, differentiate that relation and recover the original equation on the required branch. A correct constant alone does not prove a correct solution when a sign, denominator or time bound has been lost.
Pause and explain
Does checking y(0) alone prove that a formula solves a differential equation?
Worked example
Solve dy/dx = x/y with y(0) = 2 and verify the result.
Show the worked solution
- Integrate y dy = x dx to obtain y² = x² + C.
- The condition gives C = 4 and selects the positive square root y = √(x² + 4).
- Its derivative is x/y and its value at 0 is 2; the radicand is positive for all real x.
Answer y = √(x² + 4), for all real x.
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 = 3y with y(0) = −2 and verify both requirements.
Give me a hint
A negative starting value is carried by the multiplicative constant.
Compare my reasoning
- Separation gives y = Ae^(3x), including the possible equilibrium A = 0.
- The condition sets A = −2, so y = −2e^(3x).
- Differentiate to obtain −6e^(3x) = 3y, and evaluate y(0) = −2.
y = −2e^(3x), for all real x.
Look for these in your work
- Retain the sign when exponentiating a logarithmic relation.
- Check both the equation and the initial value.
Solve dy/dx = 2x/y with y(1) = −3, retaining the appropriate branch.
Give me a hint
Integrate first; then use the original signed starting value.
Compare my reasoning
- Integrate y dy = 2x dx to obtain y² = 2x² + C.
- Use 9 = 2 + C, giving C = 7 and y = −√(2x² + 7).
- The derivative −2x/√(2x² + 7) equals 2x/y, and y(1) = −3.
y = −√(2x² + 7), for all real x.
Look for these in your work
- Use the non-zero starting input when determining C.
- Keep the negative root and verify the original denominator stays non-zero.
A supplied positive-amount model is dy/dt = −y² with y(0) = 2, where t is elapsed hours. Find y(t) and the time at which the amount first reaches 1.
Give me a hint
Integrate the reciprocal square and impose the starting value.
Compare my reasoning
- Integration gives 1/y = t + C; the condition gives C = 1/2.
- Thus y = 2/(1 + 2t) for t ≥ 0; its derivative equals −y².
- Set 2/(1 + 2t) = 1 to obtain t = 1/2 hour; the amount decreases on the forward interval.
y(t) = 2/(1 + 2t), t ≥ 0; reaches 1 at t = 1/2 h.
Look for these in your work
- Verify the initial condition and rate equation.
- Use the forward interval and solve the stated threshold rather than a different fraction.
Common mistakes
- Substitute the actual starting input, which need not be zero.
- A squared initial condition loses its sign; use it to choose the root branch.
- Do not continue an initial-value solution through a pole or forbidden denominator.
What can you explain now?
For y² = x² + 4 and y(0) = −2, which branch is selected?
Compare with the explanation
y = −√(x² + 4).
The positive branch has the wrong initial sign even though it satisfies the same squared relation.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Use a non-zero-input or signed initial condition, choose a branch and verify the function against its derivative law and interval of definition.
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.