1.3 Separable equations
Theorem 1.3.1. Let \(g\) be continuous on an interval \(J\) and \(h\) continuous and non-zero on an interval \(K\). Then a differentiable function \(y\colon I \to K\) solves \(y' = g(t)h(y)\) if and only if \[ H\bigl (y(t)\bigr ) = G(t) + C \quad \text {on } I, \qquad \text {where } H(y) = \int \frac {dy}{h(y)}, \quad G(t) = \int g(t)\,dt, \] for some constant \(C\). Since \(H\) is strictly monotone on \(K\), the relation can be solved, \(y(t) = H^{-1}\bigl (G(t) + C\bigr )\), wherever \(G(t) + C\) lies in the range of \(H\).
Proof. \(H' = 1/h\) is continuous and of one sign on \(K\), so \(H\) is strictly monotone and invertible. By the chain rule, \(\frac {d}{dt}H(y(t)) = y'(t)/h(y(t))\), which equals \(g(t) = G'(t)\) for all \(t\) if and only if \(y' = gh(y)\). Two functions with the same derivative on an interval differ by a constant. □
The hypothesis \(h \neq 0\) matters. Every zero \(y^*\) of \(h\) gives a constant solution \(y \equiv y^*\), an equilibrium, which the formal method of dividing by \(h(y)\) loses; the separated solutions live in the intervals between the zeros of \(h\). Whether a separated solution can reach an equilibrium in finite time is a question about uniqueness, and the answer can be yes.
Example 1.3.2. The problem \(y' = 3y^{2/3}\), \(y(0) = 0\) has the solution \(y \equiv 0\). Separating, \(y^{1/3} = t + C\), gives \(y = t^3\), which is also a solution with \(y(0) = 0\). Worse, for any \(a \leq 0 \leq b\) the function equal to \((t - a)^3\) for \(t < a\), to \(0\) on \([a, b]\), and to \((t - b)^3\) for \(t > b\) is a solution: there are infinitely many. The right-hand side \(3y^{2/3}\) is continuous but its derivative is infinite at \(y = 0\). The next section shows that this is the only way uniqueness can fail for continuous \(f\).
Example 1.3.3. For the logistic equation \(y' = ry(1 - y/K)\) with \(0 < y_0 < K\), separation with partial fractions, \(\int \bigl (\frac 1y + \frac {1}{K - y}\bigr )dy = rt + C\), gives \(\log \frac {y}{K - y} = rt + C\), and with the initial condition \[ y(t) = \frac {K}{1 + (K/y_0 - 1)e^{-rt}} . \] The solution rises along an S-shaped curve from \(y_0\) towards \(K\), with its fastest growth when \(y = K/2\), at time \(t^* = \frac 1r\log (K/y_0 - 1)\). The same formula holds for \(y_0 > K\), when the solution decreases to \(K\), and the equilibria \(0\) and \(K\) are the solutions lost by separation.
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