Order, degree and linearity
A differential equation has derivatives of an unknown function. Its order is the highest derivative, and its degree is the power of that highest derivative. It is linear if the unknown function and its derivatives appear only to the first power and are not multiplied together.
| Equation | Order | Degree | Linear? |
|---|---|---|---|
| dy/dx = 3x | 1 | 1 | Yes |
| d²y/dx² + 4y = 0 | 2 | 1 | Yes |
| (dy/dx)² + y = x | 1 | 2 | No |
| y dy/dx = x | 1 | 1 | No (y multiplies dy/dx) |
First-order equations
Separable: dy/dx = f(x)g(y). Write dy/g(y) = f(x) dx and integrate both sides. Linear first-order: dy/dx + P(x)y = Q(x). Multiply by the integrating factor e^(∫P dx) and integrate.
Solve dy/dx = 2xy with y = 1 when x = 0.
- Separate the variables.
- Integrate.
- y = 1 at x = 0 gives C = 0.
Answer: The solution is y = e^(x²).
Solve dy/dx + y = e^(−x). The integrating factor is eˣ.
- Multiply by eˣ.
- Integrate.
Answer: The general solution is y = (x + C)e⁻ˣ.
Second-order linear equations with constant coefficients
For ay″ + by′ + cy = 0, try y = e^(mx). This gives the auxiliary equation am² + bm + c = 0. Two real roots m₁, m₂ give y = Ae^(m₁x) + Be^(m₂x); a repeated root m gives y = (A + Bx)e^(mx); complex roots p ± qi give y = e^(px)(A cos qx + B sin qx). For a non-homogeneous equation, the solution is the complementary function plus a particular integral.
Solve y″ − 3y′ + 2y = 0.
- Auxiliary equation.
Answer: The general solution is y = Aeˣ + Be²ˣ.
The solution y = A cos kx oscillates. Choose A, k and x.
Key terms
- Order
- The highest derivative in the equation.
- Degree
- The power of the highest derivative.
- Integrating factor
- A multiplier that makes a linear equation integrable.
- Auxiliary equation
- The quadratic in m that comes from trying e^(mx).
Summary
- Order is the highest derivative; degree is its power.
- Separable equations: separate and integrate.
- Linear first-order: use the integrating factor.
- Second-order: solve the auxiliary equation.