What Does "Separable" Mean?
A differential equation connects a function \( y(x) \) to its rate of change \( \frac{dy}{dx} \). A separable equation is one where the right-hand side can be factored into a piece that only involves \( x \) multiplied by a piece that only involves \( y \).
An ODE is separable if it can be written as:
The \(x\)-stuff and \(y\)-stuff can be completely divorced — no mixing, no cross-terms.
Imagine you're trying to figure out how fast a population grows. The rate depends on two things: how many creatures are already there (a \(y\) thing) AND how favorable the environment is at time \(t\) (an \(x\) thing). If those two influences just multiply together rather than tangle up in some complicated way, the equation is separable.
How to Spot One
Try to factor \( f(x,y) \) on the right side. If you can split it into \( g(x) \cdot h(y) \), you're done.
| Equation | Separable? | Factored Form |
|---|---|---|
| \(\dfrac{dy}{dx} = xy\) | ✓ Yes | \(g(x)=x,\quad h(y)=y\) |
| \(\dfrac{dy}{dx} = x + y\) | ✗ No | Can't factor — terms add, not multiply |
| \(\dfrac{dy}{dx} = \dfrac{x^2}{y^2}\) | ✓ Yes | \(g(x)=x^2,\quad h(y)=\frac{1}{y^2}\) |
| \(\dfrac{dy}{dx} = e^{x+y}\) | ✓ Yes | \(e^{x+y} = e^x \cdot e^y\) |
| \(\dfrac{dy}{dx} = \sin(xy)\) | ✗ No | No clean factoring |
The exponential \( e^{x+y} = e^x \cdot e^y \) is a classic disguise. Always try to use exponent rules to split them.
The Separation of Variables Method
Here is the complete procedure. We are going to understand why each step is legal, not just memorize the moves.
We have: \(\quad \dfrac{dy}{dx} = g(x)\,h(y)\)
The Four Moves
Start with the equation:
$$\frac{dy}{dx} = g(x)\,h(y)$$Divide both sides by \(h(y)\) (assuming \(h(y) \neq 0\)):
$$\frac{1}{h(y)}\frac{dy}{dx} = g(x)$$Integrate both sides with respect to \(x\):
$$\int \frac{1}{h(y)}\frac{dy}{dx}\,dx = \int g(x)\,dx$$Apply the substitution \( dy = \frac{dy}{dx}dx \) to the left side:
$$\int \frac{1}{h(y)}\,dy = \int g(x)\,dx$$Think of \(\frac{dy}{dx}\) as a fraction (this is informal but captures the idea). When you write \(\frac{dy}{dx} \cdot dx\), the \(dx\)'s "cancel" and you're left with \(dy\). More rigorously: by the substitution rule for integrals, if \(y\) is a function of \(x\), then \(\int f(y)\,\frac{dy}{dx}\,dx = \int f(y)\,dy\). This is just the Chain Rule reversed.
Each side is now a pure, single-variable integral. Evaluate them both, combine constants, done.
The "Leibniz Shortcut" Notation
In practice, people write the separation more suggestively. Starting from step ①:
It looks like we "multiplied \(dx\) to the right side." That's not exactly what's happening (the rigorous version is the substitution above), but it gives the correct answer and is the standard notation you'll see everywhere.
Why Is This Actually Valid?
Let's go deeper and prove the separation step rigorously using the Chain Rule. This is where the real understanding lives.
Suppose \(\frac{dy}{dx} = g(x)h(y)\) and let \(H(y)\) be an antiderivative of \(\frac{1}{h(y)}\), i.e., \(H'(y) = \frac{1}{h(y)}\). Let \(G(x)\) be an antiderivative of \(g(x)\). Then:
Define \( F(x) = H(y(x)) \) — the composite function where we plug our solution \(y(x)\) into \(H\).
Differentiate \(F\) with respect to \(x\) using the Chain Rule:
$$\frac{d}{dx}H(y(x)) = H'(y(x))\cdot y'(x) = \frac{1}{h(y)}\cdot\frac{dy}{dx}$$Substitute the ODE \(\frac{dy}{dx} = g(x)h(y)\) into the expression:
$$\frac{d}{dx}H(y(x)) = \frac{1}{h(y)}\cdot g(x)h(y) = g(x)$$We now have \(\frac{d}{dx}H(y(x)) = \frac{d}{dx}G(x)\). If two functions have the same derivative everywhere, they differ by at most a constant:
$$H(y(x)) = G(x) + C$$Translating back to integral notation:
$$\int \frac{dy}{h(y)} = \int g(x)\,dx + C$$The Chain Rule is the engine powering this whole method. When we wrote \(\frac{1}{h(y)}\frac{dy}{dx}\), we were actually constructing the exact derivative of the composite function \(H(y(x))\). The ODE then tells us that derivative equals \(g(x)\), which means \(H(y(x))\) and \(G(x)\) track each other — they're the same function up to a constant.
Implicit vs Explicit Solutions
After separating and integrating, you get an equation that implicitly defines \(y\) as a function of \(x\). Sometimes you can solve for \(y\) explicitly; sometimes you can't (or it's too messy to bother).
A relation \(F(x,y) = C\) that the solution curve satisfies — we haven't isolated \(y\).
Both \(y = \sqrt{C-x^2}\) and \(y = -\sqrt{C-x^2}\) satisfy this.
The function \(y = f(x)\) written out directly — \(y\) is isolated on the left.
\(y\) is a function of \(x\) alone, no ambiguity.
Implicit is like saying "the answer is somewhere on this circle." Explicit is like saying "the answer is this specific point on the circle." Both describe the geometry, but explicit is more direct. You can always check an implicit solution by differentiating implicitly.
When Can You Solve for \(y\) Explicitly?
After integration, you get something like \( H(y) = G(x) + C \). You can solve explicitly when \(H\) is easily invertible. Common cases:
| After Integration | Explicit Form | Notes |
|---|---|---|
| \(\ln|y| = G(x) + C\) | \(y = Ae^{G(x)}\) where \(A=\pm e^C\) | Exponential — always invertible |
| \(\frac{y^2}{2} = G(x) + C\) | \(y = \pm\sqrt{2G(x)+2C}\) | Two branches — need initial condition to pick one |
| \(\arctan(y) = G(x)+C\) | \(y = \tan(G(x)+C)\) | Invertible on the right domain |
| \(y^5 + y = G(x)+C\) | Not easily solvable | Leave as implicit |
The constant of integration from both sides can always be combined into one constant \(C\). You only need one arbitrary constant for a first-order ODE (one integration step). So write \(\int \frac{dy}{h(y)} = \int g(x)\,dx + C\), not \(+C_1 - C_2\).
Worked Examples, Full Detail
This models population growth, radioactive decay (with \(k<0\)), and countless other phenomena. Let's solve it from scratch.
Identify the structure: \(g(x) = k\) (constant), \(h(y) = y\).
Assume \( y \neq 0 \), divide both sides by \(y\):
$$\frac{1}{y}\,\frac{dy}{dx} = k$$Rewrite using Leibniz notation and separate:
$$\frac{dy}{y} = k\,dx$$Integrate both sides:
$$\int \frac{dy}{y} = \int k\,dx$$ $$\ln|y| = kx + C$$Exponentiate both sides to solve for \(y\) explicitly:
$$|y| = e^{kx+C} = e^C \cdot e^{kx}$$ $$y = \pm e^C \cdot e^{kx}$$Let \(A = \pm e^C\) be a new arbitrary constant (absorbing the ±). Note: \(y=0\) is also a solution (the trivial one we divided away). So the general solution is:
$$\boxed{y = Ae^{kx}}$$The ODE says: the rate of change of \(y\) is proportional to \(y\) itself. If you're big, you grow fast. If you're small, you grow slow. The only functions where "the rate of change equals a constant times the function itself" are exponentials — that's what this equation is selecting for.
Here both \(x\) and \(y\) appear, but they multiply — so it's separable.
Identify: \(g(x) = x\), \(h(y) = y^2\). Separate (assume \(y \neq 0\)):
$$\frac{dy}{y^2} = x\,dx$$ $$y^{-2}\,dy = x\,dx$$Integrate both sides:
$$\int y^{-2}\,dy = \int x\,dx$$ $$\frac{y^{-1}}{-1} = \frac{x^2}{2} + C$$ $$-\frac{1}{y} = \frac{x^2}{2} + C$$Solve explicitly for \(y\):
$$\frac{1}{y} = -\frac{x^2}{2} - C$$Let \(K = -C\) (just renaming the constant):
$$\boxed{y = \frac{1}{K - \frac{x^2}{2}}}$$We divided by \(y^2\), so we assumed \(y \neq 0\). Check: is \(y = 0\) also a solution? Plug in: \( \frac{d(0)}{dx} = x \cdot 0^2 \Rightarrow 0 = 0\). Yes! \(y = 0\) is a singular solution — it can't be obtained from the general solution by any choice of \(K\).
\(g(x) = \cos x\), \(h(y) = \frac{1}{y}\). Separate:
$$y\,dy = \cos x\,dx$$Integrate both sides:
$$\int y\,dy = \int \cos x\,dx$$ $$\frac{y^2}{2} = \sin x + C$$Implicit solution: \(\dfrac{y^2}{2} = \sin x + C\)
Explicit (two branches):
$$\boxed{y = \pm\sqrt{2\sin x + C'}} \quad (C' = 2C)$$Initial Value Problems
The general solution of a first-order ODE contains one arbitrary constant \(C\). An initial condition \(y(x_0) = y_0\) pins down exactly which solution curve you're on.
The general solution is a whole family of curves filling the plane — one for each value of \(C\). The initial condition is like dropping a pin at the point \((x_0, y_0)\): exactly one curve from the family passes through that point. That's your particular solution.
IVP Procedure
Stage 1 — Solve the ODE to get the general solution \(y = f(x, C)\).
Stage 2 — Substitute \(x = x_0\), \(y = y_0\) and solve for \(C\).
Stage 1 — General Solution
Separate: \(\quad \dfrac{dy}{y} = 2x\,dx\)
Integrate: \(\quad \ln|y| = x^2 + C\)
Exponentiate: \(\quad y = Ae^{x^2}\) where \(A = \pm e^C\)
Stage 2 — Apply Initial Condition
Plug in \(y(0) = 3\):
$$3 = A\,e^{0^2} = A\,e^0 = A \cdot 1 = A$$Particular solution:
$$\boxed{y = 3e^{x^2}}$$Differentiate: \(y' = 3 \cdot 2x \cdot e^{x^2} = 6xe^{x^2}\). Now check the ODE: \(2xy = 2x \cdot 3e^{x^2} = 6xe^{x^2}\). ✓ Check initial condition: \(y(0) = 3e^0 = 3\). ✓ Always verify both the ODE and the initial condition.
Separate: \(\quad y\,dy = x\,dx\)
Integrate both sides:
$$\frac{y^2}{2} = \frac{x^2}{2} + C$$ $$y^2 = x^2 + K \quad (K = 2C)$$Apply initial condition \(y(1) = -2\):
$$(-2)^2 = (1)^2 + K \implies 4 = 1 + K \implies K = 3$$Implicit solution: \(\quad y^2 - x^2 = 3\) — this is a hyperbola!
To find the explicit solution, we solve for \(y\):
$$y = \pm\sqrt{x^2 + 3}$$Since \(y(1) = -2 < 0\), we pick the negative branch:
$$\boxed{y = -\sqrt{x^2 + 3}}$$Traps and Pitfalls
You get ONE constant \(C\) total — not one per side. \(\int \frac{dy}{y} = \int k\,dx\) gives \(\ln|y| + C_1 = kx + C_2\), which collapses to \(\ln|y| = kx + C\) where \(C = C_2 - C_1\). Only write one \(C\).
When you divide by \(h(y)\) to separate, you implicitly assume \(h(y) \neq 0\). Always check separately: does \(h(y) = 0\) give a constant solution (equilibrium) that satisfies the ODE? If yes, add it as a singular solution.
\(\int \frac{dy}{y} = \ln|y|\), NOT \(\ln y\). When you exponentiate: \(|y| = e^C e^{kx}\), so \(y = \pm e^C e^{kx}\). The \(\pm\) is absorbed into the arbitrary constant \(A\), but skipping the absolute value can cause errors when \(y < 0\).
\(\frac{dy}{dx} = x + y\) is NOT separable. No matter how hard you try, \(x + y\) cannot be written as \(g(x) \cdot h(y)\). (It can only be written as a sum, not a product.) Trying to force separation here gives wrong answers.
Always plug your solution back into the original ODE and the initial condition. Integration errors are common and verification catches them before they propagate.
Test Your Intuition
Try each question before expanding the answer.
Separate: \(\dfrac{dy}{y} = \dfrac{dx}{x}\)
Integrate: \(\ln|y| = \ln|x| + C\)
Exponentiate: \(|y| = e^C|x|\), so \(y = Ax\) for some constant \(A\).
Initial condition: \(y(1) = 2 \Rightarrow A \cdot 1 = 2 \Rightarrow A = 2\)
Answer: \(\boxed{y = 2x}\) — a straight line through the origin!
The ODE \(\frac{dy}{dx} = \frac{y}{x}\) says "the slope at any point equals the \(y/x\) ratio," which is exactly the slope of the line from the origin to that point. So the solutions ARE lines through the origin. Pure geometric elegance.
When separating, we'd divide by \(h(y) = y(1-y)\). This is zero when \(y = 0\) or \(y = 1\). Check both:
\(y = 0\): \(\frac{d(0)}{dx} = 0\) and \(0(1-0) = 0\). ✓ Solution.
\(y = 1\): \(\frac{d(1)}{dx} = 0\) and \(1(1-1) = 0\). ✓ Solution.
Both \(y = 0\) and \(y = 1\) are equilibrium (singular) solutions — constant functions that satisfy the ODE. These are the stable points of the logistic equation. The general solution (via partial fractions) is \(y = \dfrac{1}{1 + Ae^{-x}}\), which approaches 0 as \(x \to -\infty\) and 1 as \(x \to +\infty\).
A first-order ODE involves only \(y\) and \(y'\) — one derivative. To "undo" one derivative, you integrate once. One integration produces one constant. Geometrically, specifying one piece of information (the initial value \(y(x_0) = y_0\)) is enough to uniquely select a solution from the family, which confirms there's exactly one degree of freedom — one constant. Second-order ODEs need two integrations, produce two constants, and need two initial conditions to pin them down.
A separable ODE factors as \(\frac{dy}{dx} = g(x)h(y)\). The method pulls the variables apart using the Chain Rule (working backwards), producing two independent integrals. The result is either an implicit relation \(H(y) = G(x) + C\) or an explicit function \(y = f(x)\). Adding an initial condition fixes \(C\) uniquely, selecting one curve from the family of solutions. Always check for singular solutions lost when dividing by \(h(y)\), and always verify your answer.