Matching Slope Fields, and Using Initial Conditions Honestly
- Match a slope field to its differential equation by testing structural features
- Sketch a particular solution through a given point on a slope field
- Apply an initial condition to resolve the constant, the sign, and the domain
Match by structure, not by drawing
A slope-field matching question is answered fastest by asking what dy/dx depends on, which is a structural question you can settle by looking rather than computing. If dy/dx depends only on x, then the slopes are identical along every vertical line — the field looks like columns. If it depends only on y, the slopes are identical along every horizontal line — rows. If it depends on both, neither pattern holds. Then find the zero slopes: horizontal segments occur exactly where dy/dx = 0, so a field with horizontal marks along the whole x-axis is telling you that dy/dx = 0 when y = 0, which points to something with a factor of y. Two structural checks usually eliminate three of four options.
Sketching a particular solution
Given a slope field and an initial point, a solution curve is drawn by following the segments: start at the given point and move so that the curve is always tangent to the nearby marks, extending in both directions until you leave the field. Three rules that questions test. The curve must pass through the given point — a common error is to draw a plausible curve that misses it. The curve must not cross a horizontal-slope line where the field shows zero slope everywhere along it, because such a line is itself an equilibrium solution and solutions cannot cross one another. And the sketch should extend across the full width of the field unless the curve leaves the top or bottom, since a half-drawn curve loses credit for incompleteness.
Separation of variables: the three decisions after integrating
The mechanics of separation are routine — collect y terms with dy, x terms with dx, integrate both sides, and include a single constant C. What earns and loses points is what happens next. First, resolve C with the initial condition before simplifying further, which is almost always easier than carrying C through an exponentiation. Second, decide the sign. Solving y² = something or |y| = something leaves two branches, and the initial condition selects one: if y(0) = −2, the solution is the negative branch and writing the positive root is simply wrong. Third, state the domain. A solution is valid on the largest interval containing the initial x on which it is defined and differentiable, so a solution with a factor of (4 − x²) under a root, or a denominator vanishing at x = 3, is not valid past that point. Free-response rubrics award the sign and domain decisions separately from the algebra.
Solve dy/dx = 6xy with y(0) = 4, and state the domain of the solution.
- 1.Separate: dy/y = 6x dx, valid where y ≠ 0.
- 2.Integrate both sides: ln|y| = 3x² + C.
- 3.Apply the initial condition before exponentiating. At x = 0, y = 4: ln|4| = 0 + C, so C = ln 4.
- 4.So ln|y| = 3x² + ln 4, giving |y| = e^(3x² + ln 4) = 4e^(3x²).
- 5.Choose the sign: y(0) = 4 is positive, so y = 4e^(3x²) rather than the negative branch.
- 6.Domain: e^(3x²) is defined and positive for all real x, and y is never zero, so the solution is valid on (−∞, ∞).
Two habits that lose points. Do not write ln|y| = 3x² + C and then jump to y = e^(3x²) + C — exponentiating a sum gives a product, so the constant becomes a multiplicative factor, not an additive one. And do not drop the absolute value before the initial condition has told you which branch you are on.
A slope field shows identical slopes along every horizontal line, and horizontal segments along the line y = 2. Which differential equation matches?
After separating and integrating, a student obtains ln|y| = x² + C with y(0) = 5. The correct particular solution is
Answer the 2 checkpoints as you read.
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