A differential equation is separable when you can rewrite it so that all the y terms (including dy) are on one side and all the x terms (including dx) are on the other. In formal terms, a first-order equation dy/dx = F(x, y) is separable if and only if the right side can be factored into a function of x alone multiplied by a function of y alone: dy/dx = f(x)·g(y). If that factoring isn’t possible, the equation is not separable.
This property is what allows you to split the equation in two, integrate each side independently, and arrive at a solution. It’s one of the first solving techniques taught in a differential equations course because, when it works, it reduces the problem to two straightforward integrals.
The Factoring Test
The core question is: can you express the right side of dy/dx = F(x, y) as one piece that depends only on x times another piece that depends only on y? If any term mixes x and y together in a way you can’t pull apart, the equation fails the test.
Consider these examples of equations that are separable:
- dy/dx = x²y factors as (x²)(y), cleanly separated.
- dy/dx = sin(x)cos(x) has no y at all, which is fine. You can think of the y-part as just 1.
- dy/dx = (x² + 1)eʸy factors as (x² + 1) times (eʸy).
And these are not separable:
- dy/dx = x² + y is a sum, not a product. No factoring will isolate x from y.
- dy/dx = (xy + 1)eʸy has xy + 1 in a way that can’t be split into separate x and y factors.
- dy/dx − sin(x)cos(y) = 1 becomes dy/dx = sin(x)cos(y) + 1, and that addition of 1 prevents clean factoring.
The distinction often comes down to multiplication versus addition. Products of x-only and y-only terms are separable. Sums that entangle x and y are not.
How to Solve a Separable Equation
Once you’ve confirmed an equation is separable, the solving process has three steps. Start with dy/dx = f(x)·g(y).
First, rearrange so that y and dy are on one side and x and dx are on the other. Divide both sides by g(y) and multiply both sides by dx to get (1/g(y)) dy = f(x) dx. This “separation of variables” is the move that gives the technique its name.
Second, integrate both sides independently. The left side becomes an integral in y only, the right side an integral in x only. You get H(y) = G(x) + C, where C is a single merged constant of integration. You only need one constant, not two, because combining the two constants from each integral just produces one net constant.
Third, solve for y if possible. Sometimes you can isolate y to get an explicit solution like y = some expression in x. Other times the algebra won’t cooperate, and the best you can do is leave the relationship in its implicit form, like H(y) = G(x) + C. Both forms are valid solutions.
A Quick Example
Take dy/dx = x² · y². Divide both sides by y² and multiply by dx: (1/y²) dy = x² dx. Integrate both sides: −1/y = x³/3 + C. Solving for y gives y = −3/(x³ + 3C), which you can simplify by renaming the constant. That’s your general solution.
Equilibrium Solutions You Might Miss
There’s a subtle trap in the separation process. When you divide both sides by g(y) to isolate the variables, you’re assuming g(y) ≠ 0. But any value of y that makes g(y) = 0 is actually a constant solution to the equation, called an equilibrium solution, and dividing by g(y) throws it away.
For example, with dy/dx = (x + 1)(y² − 4), setting y² − 4 = 0 gives y = 2 and y = −2. Both are constant solutions: if y starts at 2 or −2, it stays there forever regardless of x. These solutions won’t appear in the general formula you get from separating and integrating. You have to check for them separately by solving g(y) = 0.
What Happens to the Constant
One thing that trips people up is how the constant of integration behaves as you manipulate the solution algebraically. When you integrate both sides, you get a single C. But as you exponentiate, multiply, or rearrange to solve for y, the constant transforms.
Say your integration step produces ln|y| = 3tan(2x) − 6x + C₁. Exponentiating both sides gives y = e^(C₁) · e^(3tan(2x) − 6x). Since e^(C₁) is itself just an arbitrary positive constant, you rename it C₂ and write y = C₂e^(3tan(2x) − 6x). The constant changed form but still represents one degree of freedom. If you have an initial condition (a known y value at a specific x), you plug it in at this stage to pin down C₂ and get your particular solution.
Why Separable Equations Show Up Everywhere
Many natural processes happen to produce separable equations, which is part of why this technique matters beyond the classroom.
Population growth is a classic case. The simplest model says a population P grows at a rate proportional to its current size: dP/dt = kP. This is separable since the right side is a function of P alone (times 1 as the t-function). Separating and integrating gives the exponential growth formula P(t) = P₀e^(kt), where P₀ is the starting population. This describes the early growth of bacteria, algae, and cell cultures reasonably well.
A more realistic version is the logistic equation, dP/dt = r(1 − P/C)P, where C is the environment’s carrying capacity. The right side is still purely a function of P, so the equation is separable. The extra (1 − P/C) factor causes growth to slow as the population approaches the carrying capacity, producing the familiar S-shaped curve instead of unbounded exponential growth.
Cooling and heating follow the same pattern. Newton’s law of cooling says dT/dt = −k(T − Tₐ), where Tₐ is the ambient temperature. The right side depends only on T, making it separable. Dividing by (T − Tₐ) and integrating gives T(t) = Ce^(−kt) + Tₐ for an object cooling down toward room temperature. The same structure, solved the same way.
In each of these cases, the key insight is the same: the rate of change depends on the current state of the system in a way that cleanly separates from the independent variable. That factorability is what makes the equation solvable by this method.
Implicit vs. Explicit Solutions
After separating and integrating, you’ll sometimes end up with a relationship like y³/3 = x²/2 + 4x + C rather than a clean y = something formula. Whether you can isolate y depends on the algebra. A cubic like this one can be solved (take the cube root of both sides), giving y = (3x²/2 + 12x + C)^(1/3). But many equations produce relationships involving y inside logarithms, trigonometric functions, or combinations that simply can’t be untangled into y = f(x).
An implicit solution is still a complete, valid answer. It defines y as a function of x even if you can’t write it in isolated form. When you do have an initial condition, it sometimes helps you choose between multiple branches of the implicit relationship. For instance, if an implicit solution involves y², an initial condition telling you y(0) = 3 (positive) lets you take the positive square root and discard the negative one.