📄

← Module 2 Home Lesson 1 of 4 Next Lesson →

Lesson 1: Separable Equations

Estimated time: 35-40 minutes

Learning Objectives

Recognize when a first-order ODE is separable
Separate variables and integrate both sides
Solve for y explicitly when possible; otherwise leave in implicit form
Handle initial conditions to find particular solutions
Identify and account for lost solutions when dividing by expressions involving y

What Is a Separable Equation?

Separable Equation: A first-order ODE that can be written as

dy/dx = g(x) · h(y)

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

The strategy: move all y-terms to the left and all x-terms to the right, then integrate:

∫ (1/h(y)) dy = ∫ g(x) dx

Worked Examples

Example 1: Basic Separable Equation

Solve dy/dx = xy.

Step 1: Separate. dy/y = x dx (assuming y ≠ 0).

Step 2: Integrate both sides.

ln|y| = x²/2 + C

Step 3: Solve for y.

|y| = e^{x²/2 + C} = e^C · e^x²/2

y = Ae^x²/2 where A = ±e^C (or A = 0 for the lost solution y = 0).

General solution: y = Ae^x²/2, A any real constant.

Example 2: With Initial Condition

Solve dy/dx = 2xy², y(0) = 1.

Step 1: Separate. dy/y² = 2x dx.

Step 2: Integrate.

-1/y = x² + C

Step 3: Apply y(0) = 1: -1/1 = 0 + C, so C = -1.

Step 4: Solve for y.

-1/y = x² - 1 → y = 1/(1 - x²)

Particular solution: y = 1/(1 - x²), defined for |x| < 1.

Example 3: Implicit Solution

Solve dy/dx = (x + 1)/(y² + 1).

Step 1: Separate. (y² + 1) dy = (x + 1) dx.

Step 2: Integrate.

y³/3 + y = x²/2 + x + C

We cannot solve explicitly for y, so the implicit solution is:

y³/3 + y = x²/2 + x + C

Example 4: Lost Solutions

Solve dy/dx = y(1 - y).

Step 1: Separate (assuming y ≠ 0 and y ≠ 1).

dy / [y(1-y)] = dx

Step 2: Partial fractions: 1/[y(1-y)] = 1/y + 1/(1-y).

∫ [1/y + 1/(1-y)] dy = ∫ dx

ln|y| - ln|1-y| = x + C

ln|y/(1-y)| = x + C

y/(1-y) = Ae^x → y = Ae^x/(1 + Ae^x)

Lost solutions: y = 0 and y = 1 (the equilibria we divided away). Both satisfy the original DE.

Check Your Understanding

1. Solve dy/dx = 3x² y.

Answer: Separate: dy/y = 3x² dx. Integrate: ln|y| = x³ + C. So y = Ae^x³.

2. Solve dy/dx = e^x+y. (Hint: e^x+y = e^x · e^y.)

Answer: Separate: e^-y dy = e^x dx. Integrate: -e^-y = e^x + C. So e^-y = -e^x - C, or y = -ln(-e^x - C).

3. Solve dy/dx = (1 + y²), y(0) = 0.

Answer: Separate: dy/(1+y²) = dx. Integrate: arctan(y) = x + C. y(0)=0 gives C=0. So y = tan(x), valid on (-π/2, π/2).

4. Is dy/dx = x + y separable? Why or why not?

Answer: No. The right side x + y cannot be factored as g(x)h(y). This equation is linear (Module 2, Lesson 2), not separable.

5. When solving a separable equation, why must you check for lost solutions?

Answer: When you divide both sides by h(y), you tacitly assume h(y) ≠ 0. Any constant y = c where h(c) = 0 is a valid equilibrium solution that was lost in the division step. Always check these separately.

Key Takeaways

A separable ODE has the form dy/dx = g(x)h(y); separate variables and integrate.
Solutions may be explicit (y = ...) or implicit (F(x,y) = C).
Always check for lost solutions when dividing by h(y).
Apply initial conditions after integration to determine the constant C.
The domain of the solution may be restricted (e.g., y = tan x only on (-π/2, π/2)).

Next Lesson

Learn to solve first-order linear equations using integrating factors.

Lesson 2

Module Home

Back to Module 2