1 · Separable equations — the easiest win
If dy/dx = g(x)·h(y), separate: dy/h(y) = g(x)dx, integrate both sides, rejoin with a constant C. Check: dy/dx = 2xy → dy/y = 2xdx → ln|y| = x² + C → y = Ae^{x²}.
Solve dy/dx = (1+y)/(1+x), y(0)=1. Separate: dy/(1+y) = dx/(1+x) → ln|1+y| = ln|1+x| + C → 1+y = A(1+x). Using y(0)=1 gives A=2, so y = 2x+1.
2 · First-order linear with an integrating factor
For y′ + P(x)y = Q(x), multiply by μ(x) = e^{∫P(x)dx}: (μy)′ = μQ. Integrate, divide by μ.
y′ + (1/x)y = x. Here P = 1/x, μ = e^{ln|x|} = x. Then (xy)′ = x², so xy = x³/3 + C, and y = x²/3 + C/x.
3 · Constant-coefficient second order — the characteristic equation
For ay″ + by′ + cy = 0, guess y = e^{rx} and solve ar² + br + c = 0:
- Two distinct real r₁, r₂: y = C₁e^{r₁x} + C₂e^{r₂x}.
- Repeated root r: y = (C₁ + C₂x)e^{rx}.
- Complex roots α ± βi: y = e^{αx}(C₁cos(βx) + C₂sin(βx)).
y″ − 3y′ + 2y = 0. Characteristic: r² −3r + 2 = (r−1)(r−2). So y = C₁eˣ + C₂e^{2x}.
4 · Particular solutions — when the right side fights back
For ay″ + by′ + cy = Q(x) with Q polynomial/exponential/trig, guess the same family. If a guess collides with a homogeneous solution (a “conflict”), multiply by x until it separates.
y″ − 3y′ + 2y = eˣ. Since eˣ already solves the homogeneous part, try y_p = A·x·eˣ. Substituting gives A = 1, so y_p = xeˣ. General solution: y = C₁eˣ + C₂e^{2x} + xeˣ.
5 · Initial value problems
The constants C₁, C₂ are fixed by point conditions (y(0), y′(0)). Modeling problems — population decay, loan balance, spring motion — always end with “use the initial data to choose the unique solution meaningful in context.”
Linear: y′ + P(x)y = Q(x) ⇒ μ = e^{∫Pdx}, (μy)′ = μQ
Constant-coeff homogeneous: ar² + br + c = 0 → e^{rx} forms above
Conflict rule: if a guess duplicates a homogeneous term, multiply by x
IVP: constants from y(0), y′(0), …
Common mistakes
- Forgetting the +C (or +C₁, +C₂) — the general solution is a family, not one curve.
- Satisfying the ODE but not the initial conditions.
- Using the integrating factor without checking where its pieces are continuous.
- Missing the repeated-root x factor (C₁ + C₂x)e^{rx} — it’s not a typo.
- Guessing a particular solution that duplicates the homogeneous one without the x multiplier.