Skip to content
Marlbridge

AP® Calculus AB · Unit 7 · Topic 7.7

Particular Solutions from Initial Conditions and Separation of Variables: Revision Notes (Calculus AB 7.7)

One-page recap of particular solutions: finding the constant from an initial condition, choosing the sign, stating the domain, and the integral form of a solution.

Short on time? This page is the recap. For explanations, the domain graph and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.

Recap

  • A general solution is a family with one constant. An initial condition y(a) = y₀ picks the one particular solution through (a, y₀).
  • Steps: separate → integrate both sides → add one C → substitute (a, y₀) → solve for y → choose the sign → state the domain → check.
  • Substitute into the integrated equation, not into dy/dx.
  • After y² = … take the root whose sign matches y₀. After |y| = …, y keeps the sign of y₀ (it cannot pass through 0).
  • Domain: the largest open interval containing a on which the formula is defined and satisfies the equation. Watch for division by zero, square roots of negatives and logs of non-positive numbers.

Key relationships

Situation What to do
y²/2 = G(x) + C, y₀ < 0 y = −√(2G(x) + 2C)
ln|y| = G(x) + C, y₀ < 0 y = −e^C e^(G(x)), with C from the initial point
−1/y = G(x) + C y = −1/(G(x) + C); domain stops where G(x) + C = 0
dy/dx = f(x), y(a) = y₀ y = y₀ + ∫ (a to x) f(t) dt
Value of the integral form Use a calculator for the definite integral if no antiderivative is available

Assumptions behind the method

  • The solution is a differentiable function on one interval containing a.
  • The family of solutions has exactly one member through the given point.

Mistakes to avoid

  1. Putting (a, y₀) into dy/dx to “find C”. That gives only a slope.
  2. Assuming C = y₀ without substituting.
  3. Adding C after rearranging. The curve hits the point but fails the equation.
  4. Writing ± in the final answer. Choose one sign.
  5. A domain with a gap (“all x except ±1”). Give one interval.
  6. Swapping a and y₀ in the integral form.

Quick self-check

  1. Solve dy/dx = 6x², y(1) = 4. (y = 2x³ + C; 4 = 2 + C; y = 2x³ + 2)
  2. Solve dy/dx = x/y, y(0) = −2. (y² = x² + 4, and y₀ < 0, so y = −√(x² + 4), all real x)
  3. dy/dx = y², y(0) = 1 gives y = 1/(1 − x). State the domain. (x < 1, the interval containing 0)

Next: practice questions.

AP® and Advanced Placement® are trademarks registered by the College Board, which is not affiliated with, and does not endorse, this website. Marlbridge is an independent tuition provider. It is not an authorized AP course provider, an AP school or an exam centre, and these resources are not official College Board materials. Reading them or doing simulations does not meet the laboratory requirement of an AP science course.