Separable Equations: When Variables Can Be Pulled Apart

Separable Equations: When Variables Can Be Pulled Apart | Ideasthesia

Separation of variables is the first real technique you learn for solving differential equations. It's elegant, intuitive, and shockingly powerful.

The idea: if the equation can be factored so all the y terms are on one side and all the x terms on the other, you can integrate both sides independently.

It's almost too simple. But it works for a huge class of equations.

What Makes an Equation Separable

A first-order ODE is separable if you can write it as:

dy/dx = g(x)h(y)

The right side factors into a function of x alone times a function of y alone.

Examples of separable equations:

dy/dx = xy (separable: x times y)

dy/dx = e^(x-y) (separable: e^x · e^(-y))

dy/dx = y²/x (separable: (1/x) · y²)

Examples of non-separable equations:

dy/dx = x + y (cannot factor into g(x)·h(y))

dy/dx = sin(x + y) (the argument mixes x and y)

Separability is structural. Either the equation factors, or it doesn't.

The Technique

Start with dy/dx = g(x)h(y).

Step 1: Rewrite using differentials.

dy = g(x)h(y)dx

Step 2: Separate variables (all y on left, all x on right).

dy/h(y) = g(x)dx

Step 3: Integrate both sides.

∫dy/h(y) = ∫g(x)dx + C

Step 4: Solve for y (if possible).

That's it. The method is mechanical.

Example 1: Exponential Growth

Equation: dy/dx = ky (k constant)

This is separable: dy/dx = k · y

Separate: dy/y = k dx

Integrate: ∫dy/y = ∫k dx

ln|y| = kx + C₁

Exponentiate: |y| = e^(kx + C₁) = e^(C₁)e^(kx)

Let C = ±e^(C₁) (absorb sign and constant):

y = Ce^(kx)

This is exponential growth (k > 0) or decay (k < 0).

Physical interpretation: Any process where the rate of change is proportional to current amount. Radioactive decay, population growth, compound interest.

Example 2: Logistic Equation

Equation: dy/dx = y(1 - y)

This is separable: dy/dx = 1 · y(1 - y)

Separate: dy/(y(1 - y)) = dx

The left side needs partial fractions:

1/(y(1 - y)) = A/y + B/(1 - y)

Solve: 1 = A(1 - y) + By

Set y = 0: 1 = A, so A = 1.

Set y = 1: 1 = B, so B = 1.

Thus: 1/(y(1 - y)) = 1/y + 1/(1 - y)

Integrate: ∫(1/y + 1/(1 - y))dy = ∫dx

ln|y| - ln|1 - y| = x + C₁

ln|y/(1 - y)| = x + C₁

Exponentiate: y/(1 - y) = Ce^x (where C = ±e^(C₁))

Solve for y:

y = Ce^x(1 - y)

y = Ce^x - Ce^x y

y + Ce^x y = Ce^x

y(1 + Ce^x) = Ce^x

y = Ce^x/(1 + Ce^x)

Divide numerator and denominator by e^x:

y = C/(Ce^(-x) + 1)

Rewrite: y = 1/(1 + Ae^(-x)) (where A = 1/C)

This is the logistic function—sigmoid curve approaching 1.

Physical interpretation: Population growth with carrying capacity. Growth slows as population approaches maximum.

Example 3: Newton's Law of Cooling

Equation: dT/dt = -k(T - T_a) (T_a = ambient temperature)

Rewrite: dT/dt = -k·(T - T_a)

This is separable.

Separate: dT/(T - T_a) = -k dt

Integrate: ∫dT/(T - T_a) = ∫-k dt

ln|T - T_a| = -kt + C₁

Exponentiate: T - T_a = Ce^(-kt) (C = ±e^(C₁))

T = T_a + Ce^(-kt)

Apply initial condition T(0) = T₀:

T₀ = T_a + C, so C = T₀ - T_a

Solution: T = T_a + (T₀ - T_a)e^(-kt)

Physical interpretation: Object's temperature approaches ambient exponentially. Hot coffee cools toward room temperature.

Example 4: Mixing Problem

A tank holds 100 liters of water with 10 kg of salt dissolved. Pure water flows in at 5 L/min, and the mixture (kept uniform by stirring) flows out at 5 L/min. How does salt concentration change?

Let Q(t) = amount of salt at time t.

Salt in: 0 kg/min (pure water)

Salt out: (Q/100) · 5 kg/min (concentration Q/100, flow rate 5)

Differential equation: dQ/dt = 0 - 5Q/100 = -Q/20

This is separable: dQ/dt = -1/20 · Q

Separate: dQ/Q = -dt/20

Integrate: ln|Q| = -t/20 + C₁

Q = Ce^(-t/20)

Initial condition: Q(0) = 10, so C = 10.

Solution: Q = 10e^(-t/20)

The salt decays exponentially with time constant 20 minutes.

When Separation Fails

Not all equations are separable.

dy/dx = x + y cannot be written as g(x)h(y). The sum x + y doesn't factor.

For non-separable first-order equations, you need other techniques:

Integrating factor (for linear equations)
Exact equations
Substitution methods
Numerical approximation

Separation is powerful but limited.

The Constant of Integration

Notice every solution has an arbitrary constant C. This is the general solution—a family of curves.

To get a particular solution, apply an initial condition: y(x₀) = y₀.

Example: dy/dx = 2x with y(0) = 3

General solution: y = x² + C

Apply initial condition: 3 = 0² + C, so C = 3

Particular solution: y = x² + 3

The initial condition selects one curve from the family.

Losing Solutions: Singular Solutions

Be careful when dividing by h(y) during separation. If h(y) = 0 for some y, you might lose solutions.

Example: dy/dx = y²

Separate: dy/y² = dx

Integrate: -1/y = x + C

y = -1/(x + C)

But wait—check y = 0.

dy/dx = 0, and y² = 0. So y = 0 also satisfies the equation!

This is a singular solution (constant solution). It was lost when we divided by y².

Always check: if h(y) = 0, does that constant function satisfy the original equation? If yes, include it.

Implicit Solutions

Sometimes you can't solve for y explicitly.

Example: dy/dx = -x/y

Separate: y dy = -x dx

Integrate: y²/2 = -x²/2 + C

Simplify: x² + y² = 2C

This is a circle. You can't write y = f(x) as a single-valued function (a circle isn't a function of x).

The solution is implicit: F(x, y) = 0.

Implicit solutions are valid. They satisfy the differential equation even if you can't isolate y.

Separation with Substitution

Sometimes equations aren't obviously separable but become so after substitution.

Homogeneous equations: Form dy/dx = f(y/x)

Substitute v = y/x, so y = vx and dy/dx = v + x dv/dx.

The equation becomes separable in v and x.

Example: dy/dx = (y/x) + 1

This is homogeneous (depends on y/x).

Let v = y/x, so dy/dx = v + x dv/dx.

Substitute: v + x dv/dx = v + 1

x dv/dx = 1

Separate: dv = dx/x

Integrate: v = ln|x| + C

Back-substitute: y/x = ln|x| + C

y = x ln|x| + Cx

Substitution transforms non-separable into separable.

Applications of Separable Equations

1. Exponential growth/decay: dy/dt = ky

All processes where rate proportional to amount.

2. Logistic growth: dy/dt = ry(1 - y/K)

Population with carrying capacity, epidemic models.

3. Newton's cooling: dT/dt = -k(T - T_a)

Temperature equilibration.

4. Free fall with air resistance: dv/dt = g - kv

Velocity under gravity and drag.

5. Chemical kinetics: dC/dt = -kC^n

Reaction rates (first-order if n = 1, second-order if n = 2).

6. Torricelli's law: dh/dt = -k√h

Draining tank—height decreases as square root.

All separable. All solved by the same technique.

Why Separation Works

Mathematically, separation of variables is justified by the chain rule.

If dy/dx = g(x)h(y), then:

dy = g(x)h(y)dx

Divide by h(y):

dy/h(y) = g(x)dx

Integrate both sides with respect to their respective variables:

∫dy/h(y) = ∫g(x)dx

The left side integrates with respect to y, the right with respect to x. The equality holds because dy and dx are differentials satisfying the original equation.

It's a formal manipulation that works because derivatives and integrals are inverse operations.

Practical Tips

Tip 1: Always check for singular solutions (where you divided by zero).

Tip 2: Don't forget the absolute value in logarithms: ∫dy/y = ln|y|, not ln(y).

Tip 3: You can absorb signs into the constant. If ln|y| = x + C₁, then y = ±e^(C₁)e^x = Ce^x where C can be positive or negative.

Tip 4: Check your solution by differentiating and verifying it satisfies the original equation.

Tip 5: If the equation isn't obviously separable, try algebraic manipulation or substitution.

When to Use Separation

Use separation of variables if:

The equation is first-order
The right side factors into g(x)h(y)
You can integrate both ∫dx/g(x) and ∫dy/h(y)

If any condition fails, use a different method.

Limitations

Separation requires:

Separability: The equation must factor
Integrable factors: Both integrals must be computable

If ∫dy/h(y) or ∫g(x)dx can't be expressed in elementary functions, you're stuck.

Example: dy/dx = e^(x²)y

Separable: dy/y = e^(x²)dx

But ∫e^(x²)dx has no elementary antiderivative. You'd need numerical methods or special functions (error function).

Separation is powerful, not omnipotent.

Next Steps

Separation of variables handles a large class of first-order ODEs. But not all first-order equations are separable.

For linear first-order equations (dy/dx + P(x)y = Q(x)), you need a different technique: the integrating factor method.

We'll cover that in linear first-order ODEs.

Part 4 of the Differential Equations series.

Previous: First-Order ODEs: The Simplest Differential Equations Next: Linear First-Order: The Integrating Factor Method