How does an initial condition select one solution?
You will be able to: Use a starting value to determine a constant and verify the resulting solution.
How does an initial condition select one solution?
A slope rule is like a set of directions with no starting address. For y′=2xy, choosing y(0)=2 selects one member of the family Ae^(x²).
A useful starting point: Which solutions can disappear when you divide? →
Words and symbols before equations
- General family
- Solutions containing an arbitrary constant.
- Particular solution
- A specific function satisfying the starting data.
- Initial condition y(a)=b
- The solution must pass through (a,b).
- Verification
- Check both the derivative rule and the initial value.
What this picture assumes
Original model; rounded readouts and finite sampled graphs. y′=2xy, y(a)=b. Exact solution be^(x²−a²) with no real-domain restriction. The orange point is(a,b); displayed window[−1,1].
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=2e^(x²−0), passing through (0,2). Differentiating gives y′=2xy.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
First derive or identify the solution family. Then substitute both coordinates of the initial condition into that family; do not put the given y-value in place of the derivative.
For y=Ae^(x²) and y(0)=2, 2=Ae^0, so A=2. The solution is y=2e^(x²). It is defined for all real x.
If instead y(1)=2, then 2=Ae and A=2/e. The starting x-coordinate matters, even when the y-coordinate is unchanged.
For an equation y′=f(x), another exact representation is y(x)=b+∫ₐˣf(t)dt when f is continuous. The integral is zero at x=a and differentiates to f(x), so it automatically enforces both requirements.
A worked example, step by step
Solve y′=2xy with y(1)=3.
- Separate and integrate to get y=Ae^(x²), including the zero equilibrium.
- Substitute (1,3): 3=Ae, hence A=3/e.
- Write y=3e^(x²−1), defined for every real x.
- Differentiate to y′=6xe^(x²−1)=2xy and check y(1)=3.
Use the actual initial x-coordinate. A constant determined from the wrong point solves a different problem.
If y′=f(x), why add the initial amount to the integral?
Compare with an explanation
The integral measures change since a; b supplies the amount already present.
Predict. Change one thing. Explain.
Change the starting coordinate a and value b. Locate the marked initial point. Explain why the curve passes through it and how a changes the multiplicative constant.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
y=2e^(x²−0), passing through (0,2). Differentiating gives y′=2xy.
Derivation and check
For y≠0: dy/y=2x dx.
ln|y|=x²+C; y=Ae^(x²).
The chosen coefficient A=2.
All real x are valid; the displayed window is finite.
Original model; rounded readouts and finite sampled graphs. y′=2xy, y(a)=b. Exact solution be^(x²−a²) with no real-domain restriction. The orange point is(a,b); displayed window[−1,1].
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 questionGive an exact integral expression for y′=e^(−x²), y(2)=5, then verify both requirements without finding an elementary antiderivative.
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Compare with the answer and four-point rubric
- 1 point: Write y(x)=5+∫₂ˣe^(−t²)dt.
- 1 point: The integrand is continuous for all real t.
- 1 point: FTC gives y′(x)=e^(−x²).
- 1 point: At x=2 the integral is zero, giving y(2)=5.
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 1When do you use the initial condition?
To choose the arbitrary constant or anchor an accumulation integral.
RECALL 2What does ∫ₐˣf measure?
Signed change from a to x.
RECALL 3Must a solution have an elementary closed form?
No; a definite-integral expression can specify it exactly.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How does an initial condition select one solution?
- For y′=2xy and y(a)=b: y=be^(x²−a²).
- For y′=f(x): y=b+∫ₐˣf(t)dt.
Remember: Use the actual initial x-coordinate. A constant determined from the wrong point solves a different problem.
Conditions: Original model; rounded readouts and finite sampled graphs. y′=2xy, y(a)=b. Exact solution be^(x²−a²) with no real-domain restriction. The orange point is(a,b); displayed window[−1,1].
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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