For y′=2y and y(0)=0, the constant function y=0 is a solution. Can another solution leave zero later? For this equation, no. Now replace the right-hand side by 2√∣y∣ and ask the same question. The slope is still zero at the origin, but the answer changes. The distinction concerns how sensitively the rate changes with the state near zero.
First verify y=0 for both equations. Verification supplies one solution; it cannot by itself prove that there are no others.
Predict
On t≥0, try y=t² in y′=2√∣y∣. Its derivative is 2t, and the right-hand side is also 2t. Both this function and the zero function begin at zero. Predict whether one can wait at zero until a later time a and then follow (t−a)². Pay particular attention to the derivative where the two pieces meet.
Experiment
Choose a=1. Define y=0 for t≤1 and y=(t−1)² for t≥1. Compare its values with the solution that departs immediately.
Time
Immediate departure t²
Delayed departure
Delayed derivative
0
0
0
0
1
1
0
0
2
4
1
2
At the join both one-sided derivatives are zero, so the delayed function is differentiable there and satisfies the equation. The construction works for every nonnegative a. Draw two of these curves on the same axes and mark their shared starting point. Multiple solutions are an actual mathematical feature here, not numerical error.
Understand
A useful sufficient theorem concerns y′=f(t,y) with initial data y(t₀)=y₀. If f and its partial derivative with respect to y are continuous on a rectangle surrounding the initial point, a solution exists and is unique on some interval around t₀. The conclusion is local. It does not promise existence for all future times or provide a large interval automatically.
More general uniqueness results use a local Lipschitz condition in y: nearby rates cannot differ by more than a constant times the state difference, uniformly in a suitable time neighborhood. A continuous partial derivative is a convenient sufficient way to obtain this bound. For 2√∣y∣ near zero, the difference quotient relative to its value at zero grows without bound. Our explicit competing solutions demonstrate nonuniqueness.
Failure of a sufficient hypothesis does not alone prove failure of uniqueness. It tells you this theorem cannot settle the case. Here the counterexample functions do the decisive work.
Even smooth right-hand sides can produce finite-time blowup. For y′=y² with y(0)=1, separation gives y=1/(1−t). The right-hand side f=y² is smooth everywhere, so local uniqueness holds. Yet the selected solution becomes unbounded as t approaches one from below. Smoothness of the equation is not a bound on the trajectory.
Apply
Repeat the blowup calculation with y(0)=1/2. Obtain y=1/(2−t), check its derivative, and identify the forward blowup time. Explain why changing the initial value changes the available interval even though the equation is unchanged.
In a written explanation, distinguish three questions: does at least one solution exist, is it unique near the initial point, and how far can it continue? For the square-root example, use explicit formulas to answer the uniqueness question. For the quadratic example, use the singular denominator to answer the continuation question. Neither a smooth sketch nor a solver returning values answers all three.
Continue
Use these distinctions when separating variables next. Algebraic steps may discard equilibria or conceal domain restrictions, so retain the original equation and initial point as independent checks on every derived formula.
Apply · explain · revise
Practice and reasoning
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For y′=2y with y(0)=0, what is y(1)?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For y′=2y², y(0)=1/3, at what positive time does the solution blow up?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
A standard sufficient local uniqueness condition near (t₀,y₀) is:
For t≥0, show that y=0 and y=t² both solve y′=2√|y| with y(0)=0. What hypothesis fails near y=0?
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Give a delayed solution of y′=2√|y| on t≥0 with y(0)=0 that stays zero until a chosen a>0.
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Why does smooth f(t,y)=y² not imply a solution through (0,1) exists for all positive time?
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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.
