In y′=2ty, the rate is a product of a time factor and a state factor. Start with y(0)=3. At time zero the slope vanishes, but the function need not be constant because the time factor changes. Predict whether the state is increasing for positive time while it remains positive. Then find a formula by separating the two dependencies.
Before dividing by y, record that y=0 is also a solution of the original equation. The initial condition in this example excludes it, but the full equation does not.
Predict
After dividing, integration produces ln∣y∣, not ln y without qualification. Predict which sign branch the initial value selects. Also predict whether the solution at negative time has the same value as at the corresponding positive time. The eventual formula should explain both answers and remain consistent with the rate equation.
Experiment
On an interval where y is nonzero, write y′/y=2t. Integration gives ln∣y∣=t²+C. Exponentiating and absorbing the fixed sign yields y=Ae^(t²). The initial value makes A=3. The table uses symbolic times so the resulting values can be exact.
Time
Exponent t²
State 3e^(t²)
0
0
3
√(ln 2)
ln 2
6
−√(ln 2)
ln 2
6
Differentiate the final formula: y′=6t e^(t²)=2t y. This verifies the equation and shows why a positive state can decrease for negative time and increase for positive time. Repair the claim “positive y always means positive rate” by naming the missing factor.
Understand
For y′=g(t)h(y), separation on a region with h(y) nonzero gives an antiderivative relation ∫dy/h(y)=∫g(t)dt+C. This notation encodes the chain rule: differentiating the left antiderivative along y(t) gives y′/h(y). Every root of h must be checked separately as a constant solution before division.
For y′=y(1−y), division by y(1−y) loses the equilibria zero and one. Recovering a family after integration does not justify forgetting those omitted solutions. A solution selected by a non-equilibrium initial value may have additional interval restrictions.
Consider y′=y², y(0)=1/3. For nonzero y, integration gives −1/y=t+C. Initial data give C=−3, so y=1/(3−t). At t=1 the value is 1/2. Its maximal interval containing zero ends at three, where the expression becomes unbounded. The zero solution remains separate.
Separation can also produce implicit formulas and branch choices. For y′=t/y with y(0)=2, integration gives y²=t²+4. The initial condition selects the positive square root. Because that root never vanishes, it solves the original equation for every real t. Choosing the negative root would violate the initial condition; switching roots would destroy continuity.
Apply
Write a four-line audit beside a separated solution: excluded states, integrated relation, branch selected by initial data, and interval of validity. Apply it to both examples above. State why the denominator in the differential equation matters separately from denominators introduced while solving.
Use the practice to explain a lost-equilibrium error in words, not only by appending a constant formula. A complete solution family describes all applicable cases and their domains. A formula valid on one side of a singularity does not automatically connect to a formula on the other side.
Continue
Next apply separation to proportional growth and cooling. These models are useful precisely because their assumptions are explicit; keep the algebraic solution, parameter interpretation, and evidence for the model as separate parts of your explanation.
Apply · explain · revise
Practice and reasoning
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For y′=2t y with y(0)=5, find y at t=√(ln 3).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For y′=y² with y(0)=1/3, find y(1).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
When dividing y′=y(1−y) by y(1−y), which solutions must be checked separately?
Solve y′=2t y with y(0)=3 and verify the result.
Structure your argument and preserve revisions
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Solve y′=t/y with y(0)=2 and give its real solution branch.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
A learner solves y′=y² by dividing by y² and lists only y=−1/(t+C). Explain the missing solution and the interval restriction.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
