The friendliest differential equations are the ones that come apart. A first-order ODE is
separable when its right-hand side factors into a part depending only on
x and a part depending only on y:
\frac{dy}{dx} = g(x)\,h(y).
When that happens we can shepherd all the y's to one side and all the
x's to the other, then
integrate
each side on its own. It is the most direct solution method in the whole subject, and it is the one
every other technique secretly hopes to reduce to.
Treating \tfrac{dy}{dx} as a ratio of differentials — a manoeuvre
justified by the chain rule, as the theorem notes — the recipe is
\frac{dy}{h(y)} = g(x)\,dx \quad\Longrightarrow\quad \int \frac{dy}{h(y)} = \int g(x)\,dx.
The centrepiece: solving y' = ky
One separable equation governs an astonishing share of the natural world — populations,
radioactive decay, compound interest, cooling coffee. It is
\frac{dy}{dx} = ky,
"the rate of change is proportional to the amount present." Let us solve it from scratch, line by
line. Here g(x) = k (a constant) and h(y) = y.
Step 1 — separate the variables. Divide both sides by
y and multiply by dx, collecting each letter
on its own side:
\frac{dy}{y} = k\,dx.
Step 2 — integrate both sides. The left integral is a standard logarithm; the
right is the integral of a constant:
\int \frac{dy}{y} = \int k\,dx \quad\Longrightarrow\quad \ln|y| = kx + C_1.
Step 3 — exponentiate to free y. Apply
e^{(\cdot)} to both sides, and split the exponential of a sum:
|y| = e^{kx + C_1} = e^{C_1}\,e^{kx}.
Step 4 — absorb the constants. The factor
e^{C_1} is just some positive constant; dropping the absolute value lets
it carry a sign, and the case y \equiv 0 (lost when we divided by
y) is recovered by allowing the constant to be zero. Write the single
constant as C = \pm e^{C_1}:
\boxed{\,y = C\,e^{kx}\,.}
Step 5 — read off the meaning. Setting x = 0 gives
y(0) = C, so C is the initial amount. With
k > 0 this is exponential growth; with
k < 0, exponential decay. The proportional-rate law and
the exponential function are two faces of the same fact.
A second example, with an initial condition
Separable equations need not be linear. Solve the IVP
\frac{dy}{dx} = \frac{x}{y}, \qquad y(0) = 2.
Step 1 — separate. Here g(x) = x and
1/h(y) = 1/(1/y) = y, so bring the y across:
y\,dy = x\,dx.
Step 2 — integrate both sides.
\int y\,dy = \int x\,dx \quad\Longrightarrow\quad \frac{y^2}{2} = \frac{x^2}{2} + C_1.
Step 3 — tidy into an implicit relation. Multiply by 2 and rename the constant
C = 2C_1:
y^2 = x^2 + C.
Step 4 — apply the initial condition
y(0) = 2. Substitute x = 0, y = 2:
4 = 0 + C, so C = 4.
Step 5 — solve for y. Take the root, choosing the
positive branch because y(0) = 2 > 0:
y = \sqrt{x^2 + 4}.
A quick check: y' = x/\sqrt{x^2+4} = x/y, and
y(0) = 2. Both the equation and the initial condition hold. ✓
If a first-order ODE has the form
\dfrac{dy}{dx} = g(x)\,h(y) with h(y) \ne 0,
then it is solved by separating and integrating:
\int \frac{dy}{h(y)} = \int g(x)\,dx + C.
This is legitimate because, by the chain rule, if H' = 1/h and
G' = g then \tfrac{d}{dx}H(y(x)) = \tfrac{1}{h(y)}y' = g(x)
= G'(x), so H(y) = G(x) + C. Two riders:
-
the special case \tfrac{dy}{dx} = ky has solution
y = Ce^{kx} — exponential growth (k>0) or
decay (k<0), with C = y(0);
-
any constant y \equiv y_* with h(y_*) = 0
is an equilibrium solution, lost in the division by
h(y) and must be reinstated.
Pure exponential growth y' = ky is a fantasy of unlimited resources —
left alone, the population reaches the moon. Reality imposes a ceiling, the
carrying capacity M, and the fix is the
logistic equation, still separable:
\frac{dy}{dx} = k\,y\left(1 - \frac{y}{M}\right).
When y is small the bracket is near 1 and
growth is nearly exponential; as y \to M the bracket
\to 0 and growth stalls. Separating and integrating (the
y-integral needs
partial fractions)
gives the S-shaped curve
y = \frac{M}{1 + A\,e^{-kx}}, \qquad A = \frac{M - y_0}{y_0}.
It rises exponentially, bends at the inflection point y = M/2, then
levels off at M — the model behind epidemics, technology adoption, and
bounded population growth alike. The constant solutions y \equiv 0 and
y \equiv M are the equilibria the theorem warned us to keep.