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Module 2 Study Guide

First-Order ODEs: Solving Techniques

Differential Equations • Learn Without Walls

1. Separable Equations

Separable Equation: A first-order ODE that can be written as dy/dx = g(x)h(y), so that all y-terms go to one side and all x-terms to the other.
dy / h(y) = g(x) dx  →  Integrate both sides

Method

  1. Rewrite as dy/dx = g(x)h(y).
  2. Separate: (1/h(y)) dy = g(x) dx.
  3. Integrate both sides.
  4. Solve for y if possible (explicit solution) or leave implicit.
  5. Check for lost solutions where h(y) = 0 (equilibria).
Example: dy/dx = xy. Separate: dy/y = x dx. Integrate: ln|y| = x²/2 + C. Solution: y = Aex²/2. Also y = 0 is an equilibrium solution.
Always check for equilibrium solutions lost during separation (dividing by h(y) = 0).

2. First-Order Linear Equations

Standard Form: y' + P(x)y = Q(x). The key tool is the integrating factor.
Integrating factor: μ(x) = e∫P(x)dx
Multiply through: (μy)' = μQ
Solution: y = (1/μ) ∫ μQ dx + C/μ

Method

  1. Write in standard form y' + P(x)y = Q(x).
  2. Compute μ(x) = e∫P(x)dx.
  3. Multiply both sides by μ.
  4. Recognize left side as d/dx[μy].
  5. Integrate both sides and solve for y.
The integrating factor always works for first-order linear ODEs. It turns the left side into an exact derivative.

3. Exact Equations

Exact Equation: M(x,y)dx + N(x,y)dy = 0 is exact if ∂M/∂y = ∂N/∂x. Then there exists F(x,y) such that ∂F/∂x = M and ∂F/∂y = N, and the solution is F(x,y) = C.

Method

  1. Write as M dx + N dy = 0.
  2. Verify exactness: ∂M/∂y = ∂N/∂x.
  3. Integrate M with respect to x: F = ∫M dx + g(y).
  4. Differentiate F with respect to y and set equal to N to find g'(y).
  5. Integrate g'(y) to get g(y).
  6. Write solution as F(x,y) = C.
If not exact, sometimes an integrating factor μ(x) or μ(y) can make it exact.

Finding Integrating Factors for Non-Exact Equations

ConditionIntegrating Factor
(M_y - N_x)/N depends only on xμ(x) = e∫(M_y - N_x)/N dx
(N_x - M_y)/M depends only on yμ(y) = e∫(N_x - M_y)/M dy

4. Substitution Methods

Bernoulli Equations

Bernoulli Equation: y' + P(x)y = Q(x)yn where n ≠ 0, 1.
Substitution: v = y1-n
Transforms to linear: v' + (1-n)P(x)v = (1-n)Q(x)

Homogeneous Equations

Homogeneous (substitution sense): dy/dx = F(y/x). Every term has the same total degree.
Substitution: v = y/x, so y = vx and dy/dx = v + x(dv/dx)
Reduces to a separable equation in v and x.

Decision Flowchart

CheckIf Yes, Use...
Can you separate variables?Separable method
Is it y' + P(x)y = Q(x)?Integrating factor
Is M_y = N_x?Exact equation method
Is it y' + Py = Qyn?Bernoulli substitution
Is it dy/dx = F(y/x)?Homogeneous substitution