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.

If $f$ is continuous, $\int_0^x g(t)f(t)\,dt$ is differentiable by the fundamental theorem of calculus, so an equation "$f =$ something $+ \int_0^x\ldots$" forces $f$ itself to be differentiable — the equation bootstraps its own smoothness. Differentiating turns the integral into an ODE, which the uniqueness theorem then pins down once the initial values are known; those come from setting $x = 0$ in the original, where every $\int_0^0$ vanishes. Differentiation can lose information (constants disappear), so the final check against the original condition is part of the proof.

Step 1 — the right side is differentiable (FTC), so $f$ is, and $f'(x) = xf(x)$. Setting $x = 0$ in the original gives $f(0) = 1$.

Step 2 — integrating factor: $\left(f(x)e^{-x^2/2}\right)' = (f' - xf)e^{-x^2/2} = 0$, so $f(x)e^{-x^2/2}$ is constant, equal to its value $1$ at $x = 0$.

Step 3 — $f(x) = e^{x^2/2}$. Check: $1 + \int_0^x te^{t^2/2}\,dt = 1 + \left(e^{x^2/2} - 1\right) = e^{x^2/2}$ ✓.

How to find it. An unknown function inside $\int_0^x$ is an antiderivative in disguise; one differentiation removes it.

Step 1 — expand the integral as $x\int_0^xf(t)\,dt - \int_0^x tf(t)\,dt$; both pieces are differentiable, and by the product rule and FTC $f'(x) = 1 + \int_0^x f(t)\,dt + xf(x) - xf(x) = 1 + \int_0^x f(t)\,dt$.

Step 2 — differentiate again: $f''(x) = f(x)$.

Step 3 — initial values from the two equations at $x = 0$: $f(0) = 0$ and $f'(0) = 1$.

Step 4 — the solutions of $f'' = f$ are $ae^x + be^{-x}$; the initial values give $a + b = 0$, $a - b = 1$, so $f(x) = \tfrac{e^x - e^{-x}}{2} = \sinh x$. Check: $\int_0^x(x - t)\sinh t\,dt = \sinh x - x$ (integrate by parts), so the right side is $x + \sinh x - x = \sinh x$ ✓.

How to find it. The kernel $x - t$ is what one integration by parts produces: $\int_0^x(x - t)f(t)\,dt$ is the second antiderivative of $f$. Two differentiations undo two integrations.

Step 1 — the right side $f(1 - x)$ is differentiable, so $f'$ is, and $f''(x) = \tfrac{d}{dx}f(1 - x) = -f'(1 - x) = -f(x)$, applying the equation at the point $1 - x$.

Step 2 — so $f'' + f = 0$, whose solutions are exactly $f = A\cos x + B\sin x$ (uniqueness by the energy argument above).

Step 3 — the second-order equation has two constants; the first-order condition must remove one. Substitute: $f'(x) = -A\sin x + B\cos x$, while $f(1 - x) = (A\cos1 + B\sin1)\cos x + (A\sin1 - B\cos1)\sin x$. Matching coefficients: $B = A\cos1 + B\sin1$ and $-A = A\sin1 - B\cos1$, i.e. $B(1 - \sin1) = A\cos1$ and $B\cos1 = A(1 + \sin1)$. These are the same condition, because $(1 - \sin1)(1 + \sin1) = \cos^21$.

Step 4 — so $B = A\tfrac{1 + \sin1}{\cos1}$ and $f(x) = \tfrac{A}{\cos1}\left(\cos1\cos x + \sin x + \sin1\sin x\right) = C\left(\sin x + \cos(x - 1)\right)$ for an arbitrary constant $C$. Check: $f'(x) = C(\cos x - \sin(x - 1))$ and $f(1 - x) = C(\sin(1 - x) + \cos x)$, equal since $\sin(1 - x) = -\sin(x - 1)$ ✓.

How to find it. Applying the reflection $x \mapsto 1 - x$ twice is the identity, so differentiating twice returns to $f$ with a sign: a reflection condition hides a second-order equation, and the original condition then cuts the solution space in half.

All three rest on uniqueness. Two solutions of the same equation cannot meet (they would then coincide), so a solution that starts between two equilibria stays between them, and since $g$ keeps one sign there, it is monotone. For comparison, at a first point where $y$ would drop below $z$ the two would be equal with $y' \ge z'$; the smooth dependence of $F$ on $y$ turns that tangency into a contradiction (the careful version is Grönwall's inequality applied to $z - y$). Blow-up follows: a function forced to stay above a solution that becomes infinite at $x^*$ cannot exist beyond $x^*$.

Step 1 — equilibria: $y \equiv 0$ and $y \equiv 1$. A solution starting at $\tfrac12$ cannot reach either (uniqueness), so $0 < y < 1$ as long as it exists.

Step 2 — there $y' = y(1 - y) > 0$: $y$ is increasing and bounded, hence exists for all $x$ and has a limit $L \in (\tfrac12, 1]$ as $x \to \infty$.

Step 3 — if $L < 1$, then $y' \to L(1 - L) > 0$, so $y$ would grow without bound: contradiction. So $L = 1$.

Step 4 — solve: separate, $\tfrac{dy}{y(1 - y)} = \left(\tfrac1y + \tfrac{1}{1 - y}\right)dy = dx$, so $\ln\tfrac{y}{1 - y} = x + c$, and $y(0) = \tfrac12$ gives $c = 0$: $y = \tfrac{1}{1 + e^{-x}}$, which indeed stays in $(0, 1)$, increases, and tends to $1$.

How to find it. For an autonomous equation, draw the phase line first: the signs of $g$ between equilibria tell you where every solution goes, and the formula only confirms it.

Step 1 — $y' = x^2 + y^2 \ge 0$, so $y$ is non-decreasing and $y(x) \ge y(0) = 0$ for $x \ge 0$.

Step 2 — for $x \ge 1$: $y' \ge 1 + y^2$, i.e. $\tfrac{d}{dx}\arctan y = \tfrac{y'}{1 + y^2} \ge 1$.

Step 3 — integrate from $1$ to $x$: $\arctan y(x) \ge \arctan y(1) + (x - 1) \ge x - 1$, since $y(1) \ge 0$.

Step 4 — $\arctan$ is less than $\tfrac\pi2$, so the inequality is impossible once $x - 1 \ge \tfrac\pi2$. Hence the solution cannot exist on $[0, 1 + \tfrac\pi2]$: it becomes infinite at some $x^* < 1 + \tfrac\pi2 \approx 2.571$. $\blacksquare$ (Numerically $x^* \approx 2.003$; the comparison bound is crude but rigorous.)

How to find it. The $y^2$ term makes growth feed on itself; the $x^2$ term is only a nuisance, so bound it below by a constant on a half-line and compare with the equation $z' = 1 + z^2$, whose solution $\tan$ is known to explode.

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 $\mathbb R$."