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
- Rewrite as dy/dx = g(x)h(y).
- Separate: (1/h(y)) dy = g(x) dx.
- Integrate both sides.
- Solve for y if possible (explicit solution) or leave implicit.
- 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
- Write in standard form y' + P(x)y = Q(x).
- Compute μ(x) = e∫P(x)dx.
- Multiply both sides by μ.
- Recognize left side as d/dx[μy].
- 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
- Write as M dx + N dy = 0.
- Verify exactness: ∂M/∂y = ∂N/∂x.
- Integrate M with respect to x: F = ∫M dx + g(y).
- Differentiate F with respect to y and set equal to N to find g'(y).
- Integrate g'(y) to get g(y).
- 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
| Condition | Integrating Factor |
| (M_y - N_x)/N depends only on x | μ(x) = e∫(M_y - N_x)/N dx |
| (N_x - M_y)/M depends only on y | μ(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
| Check | If 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 |