Why must a particular solution include its interval?
You will be able to: Choose the branch and maximal interval containing the initial point.
Why must a particular solution include its interval?
A formula can work perfectly near a starting point and break down later. For y′=y², y(0)=1, the solution rises so fast that it becomes unbounded as x approaches 1.
A useful starting point: How does an initial condition select one solution? →
Words and symbols before equations
- Solution interval
- A connected interval on which the function and equation are valid.
- Singularity
- A point where the proposed formula or equation becomes undefined.
- Branch
- A connected piece of a formula restricted to a valid domain.
- Maximal interval
- The largest valid interval containing the initial point.
What this picture assumes
Original model; rounded readouts and finite sampled graphs. y′=y², y(0)=b>0. Maximal interval x<1/b; the display stops at0.85/b, before the pole. The other algebraic branch is not joined across the singularity. Vertical scale adapts.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- y=1/(1−1x). Pole x=1. Maximal interval through 0: x<1. Plot ends at x=0.85; vertical scale adapts.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Separate dy/y²=dx for nonzero y and integrate: −1/y=x+C. With y(0)=1, C=−1, giving y=1/(1−x).
The formula is undefined at x=1. The initial point is x=0, so the maximal solution interval containing it is (−∞,1). The separate branch x>1 is not a continuation through the singularity.
Differentiating gives y′=1/(1−x)²=y² on the chosen interval. As x→1− the solution grows without bound. The finite display stops before 1 and does not draw across it.
Other restrictions arise when solving implicit relations. For y²=x²+4 with y(0)=2, choose the positive square root; the negative branch fails the initial value. Verify any branch in the original equation, especially where a denominator could vanish.
A worked example, step by step
Solve y′=y² with y(0)=2 and state the maximal interval containing zero.
- Integrate after separation: −1/y=x+C.
- At (0,2), C=−1/2.
- Solve y=1/(1/2−x)=2/(1−2x).
- The pole is x=1/2, so the required interval is (−∞,1/2); derivative verification holds there.
A disconnected algebraic domain is not a single solution interval through a pole. Keep the interval containing the initial point.
Can the solution be extended through its vertical asymptote?
Compare with an explanation
No. It is not finite or defined at the pole, so it cannot be a differentiable solution there.
Predict. Change one thing. Explain.
Move the initial value b from 1 to 2. Observe how the marked pole moves. Explain why the graph stops before the pole and why no line should connect to the other branch.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
y=1/(1−1x). Pole x=1. Maximal interval through 0: x<1. Plot ends at x=0.85; vertical scale adapts.
Formula versus solution interval
Separate: −1/y=x+C.
Initial value gives C=−1.
The denominator vanishes at x=1.
The branch on the far side is not a continuation through the pole.
Original model; rounded readouts and finite sampled graphs. y′=y², y(0)=b>0. Maximal interval x<1/b; the display stops at0.85/b, before the pole. The other algebraic branch is not joined across the singularity. Vertical scale adapts.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the rate units, derivative signs, initial condition, domain or solution check. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionFor y′=x/y, y(0)=−2, derive the particular solution and explain its branch and domain.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Separate y dy=x dx and integrate y²=x²+C.
- 1 point: Substitute the initial point:4=C.
- 1 point: Choose y=−√(x²+4), because the starting value is negative.
- 1 point: This branch never vanishes and is defined for all real x; y′=−x/√(x²+4)=x/y verifies the original equation.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Which interval belongs to an IVP?
A connected valid interval containing its initial x-coordinate.
RECALL 2What chooses a square-root branch?
The initial value plus verification and domain restrictions.
RECALL 3Why split a graph at a pole?
A solution cannot pass through an undefined value.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why must a particular solution include its interval?
- y′=y², y(0)=b>0: y=b/(1−bx).
- Maximal interval containing 0: x<1/b.
Remember: A disconnected algebraic domain is not a single solution interval through a pole. Keep the interval containing the initial point.
Conditions: Original model; rounded readouts and finite sampled graphs. y′=y², y(0)=b>0. Maximal interval x<1/b; the display stops at0.85/b, before the pole. The other algebraic branch is not joined across the singularity. Vertical scale adapts.
Refresh Kid · AP Calculus BC Unit 7 · Objectives FUN-7.E · Review edition
Framework, scope and review status
Mapped to College Board CED Topic 7.7, FUN-7.E. CED and Fall 2026 clarifications checked September 17, 2026. All nine BC topics are included, with Euler’s method and logistic interpretation. The correction to FUN-7.B.2 is reflected: an equation may have infinitely many solutions. Focused explanations, examples, visual models and practice are original Refresh Kid work.
Rate and amount units, signs, initial data, lost equilibria, solution intervals and model assumptions are explicit. Slope fields and Euler polygons are finite illustrations; exact algebra supports identities and limits. The logistic explicit formula supports the visualization; required interpretation can be done from the rate law without deriving that formula. The smooth examples have unique solutions; no universal uniqueness claim is made.
Organic Chemistry Tutor video titles and destinations were checked, not the full videos. Khan Academy’s destination was checked; its JavaScript lesson contents were not fully readable by the research tool. OpenStax sections 4.1–4.4 supplied conceptual cross-checks. No provider questions, artwork or scripts were copied. No affiliation or endorsement is implied.
GitHub’s 3D website collection informed optional spatial inspection and camera controls. The original tank reuses self-hosted Three.js with its MIT license retained. The 2×2 dm base connects water depth to volume: 1 dm³=1 L. This tank uses a feedback-controlled pump, not a gravity-drain law. Its 3D height and labeled 2D graphs use the same exact exponential formula. Rotation changes only the view, and complete explanations remain available without WebGL.
Independent teacher review and student usability testing remain pending. Technical checks do not certify mathematical accuracy, full accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. This multi-unit archive is not assigned as a complete Unit 7 task.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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