Existence & Uniqueness (first order)
Differential Equations · Section 2.4 · generated practice set
When the theorem guarantees a unique solution near a point, and what its silence does and does not mean.
Practice this set → Fresh numbers on every attempt. No account needed.
Method
Two theorems, and which one applies depends on whether the equation is linear.
Linear: $y' + p(t)y = g(t)$
If $p$ and $g$ are continuous on an open interval $I$ containing $t_0$, there is exactly one solution on all of $I$. You can name the interval before solving: find where $p$ or $g$ break, and take the piece containing $t_0$.
Nonlinear: $y' = f(t,y)$
If $f$ and $\partial f/\partial y$ are continuous on a rectangle around $(t_0, y_0)$, there is a unique solution on some interval around $t_0$ — possibly much smaller, and the theorem does not tell you how small.
When the hypotheses fail, the theorem is silent. It does not prove that two solutions exist, only that nothing is guaranteed.
Definitions and theorems
Worked example
Does a unique solution of $y' = \sqrt{y}$, $y(0) = 0$ exist?
- $f(t,y) = \sqrt{y}$ is continuous for $y \ge 0$.
- $\partial f/\partial y = \frac{1}{2\sqrt y}$ is undefined at $y = 0$ — precisely the initial value.
- The theorem's hypotheses fail, so uniqueness is not guaranteed.
- In fact $y \equiv 0$ and $y = t^2/4$ both satisfy the IVP for $t \ge 0$, so uniqueness genuinely fails here.
Common mistakes
- Concluding non-existence. Failing the hypotheses means "no guarantee", not "no solution".
- Using the linear theorem's global interval on a nonlinear problem. Nonlinear gives only a local interval.
- Forgetting to check $\partial f/\partial y$. Continuity of $f$ alone gives existence but not uniqueness.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.