Differential Equations on the Putnam

The Putnam never asks you to solve a hard ODE; it asks you to recognize that a functional condition IS an ODE, solve the easy ODE, and then argue about existence, uniqueness, or a bound. $y' = y^2$ with $y(0) = 1$ gives $y = \tfrac{1}{1-x}$ — and the point of the problem is the blow-up at $x = 1$: no solution exists on all of $\mathbb R$. Likewise $f' = 1 + f^2$ forces $f = \tan$, which cannot be defined on an interval longer than $\pi$.

Multiplying by $e^{-x}$ converts a differential equation into "a derivative is zero" and a differential INEQUALITY into "a function is monotone" — the two-line arguments graders want in place of "obviously the solution is $e^x$". For second-order equations the analogous tool is an energy: for $y'' + y = 0$, $E = y^2 + y'^2$ has $E' = 2y'(y + y'') = 0$, so a solution with $y(0) = y'(0) = 0$ has $E \equiv 0$ and is identically zero — uniqueness for free.

The solution space is two-dimensional (a solution is determined by $y(0), y'(0)$ — that is the uniqueness theorem), so two independent solutions span everything. The same characteristic-root method solves linear recurrences, which is why the "Sequences" and "Differential Equations" units share a proof.

$\mu = x$: $(xy)' = x^2$, so $xy = \tfrac{x^3}{3} + C$, $C = \tfrac23$, and $y(2) = \tfrac12\left(\tfrac83 + \tfrac23\right) = \tfrac53$.

Let $v(x) = u(x)e^{-\beta x}$. Then $v' = (u' - \beta u)e^{-\beta x} \le 0$, so $v$ is non-increasing and $v(x) \le v(0) = u_0$, i.e. $u(x) \le u_0e^{\beta x}$. $\blacksquare$ Significance: Grönwall is how uniqueness of ODE solutions is actually proved (apply it to the square of the difference of two solutions), and it is the model for every "differential inequality $\Rightarrow$ growth bound" argument.

Since $y' \ge y^2 \ge 0$, $y$ is non-decreasing, so $y > 0$ wherever it exists and $\left(-\tfrac1y\right)' = \tfrac{y'}{y^2} \ge 1$. Integrating from $0$ to $x$: $-\tfrac{1}{y(x)} + \tfrac1c \ge x$, i.e. $\tfrac{1}{y(x)} \le \tfrac1c - x$, which is impossible once $x \ge \tfrac1c$ (the left side is positive). $\blacksquare$ Significance: this comparison is the standard way to show that solutions of $y' = y^2 + (\text{anything} \ge 0)$ have finite lifetime — a favorite Putnam "prove that no such function exists on all of $\mathbb R$."