What is a differential equation?
An equation involving an unknown function and its derivatives is a differential equation. Examples:
dy/dx = 2x
d²y/dx² + y = 0
x (dy/dx) + y = x²
They describe how quantities change: population growth, radioactive decay, cooling, motion under forces. In this chapter, y is a function of one variable x, so these are ordinary differential equations.
Order and degree
The order is the order of the highest derivative present.
The degree is the power of that highest derivative, once the equation is a polynomial in its derivatives. If it isn't (for example, it contains sin(dy/dx) or e^(dy/dx)), the degree is not defined.
Equation | Order | Degree |
|---|---|---|
dy/dx + y = eˣ | 1 | 1 |
(d²y/dx²)³ + (dy/dx)⁴ + y = 0 | 2 | 3 |
d³y/dx³ + 2(dy/dx)² = 0 | 3 | 1 |
dy/dx + sin(dy/dx) = 0 | 1 | not defined |
The order and degree are always positive integers (when defined).
General and particular solutions
A solution is a function that satisfies the equation.
The general solution contains arbitrary constants, as many as the order of the equation. For d²y/dx² + y = 0 it is y = a cos x + b sin x.
A particular solution is obtained by giving the constants specific values, usually from given conditions such as y(0) = 1.
Checking a solution. Is y = e^(3x) a solution of y″ − 9y = 0? y″ = 9e^(3x), so y″ − 9y = 0. ✓
Forming a differential equation
To form the differential equation of a family of curves with n arbitrary constants, differentiate n times and eliminate the constants. The result has order n.
Example. The family y = mx (lines through the origin). Then dy/dx = m, so eliminating m: x (dy/dx) = y.
Example. y = a eˣ + b e⁻ˣ. Differentiating twice: y″ = a eˣ + b e⁻ˣ = y, so y″ − y = 0.
Method 1: variables separable
If the equation can be written as dy/dx = g(x) h(y), separate the variables and integrate both sides:
∫ dy/h(y) = ∫ g(x) dx + C
Example. dy/dx = x/y. Then y dy = x dx, so y²/2 = x²/2 + C, which gives y² − x² = C′, a family of hyperbolas.
Example. dy/dx = (1 + y²)/(1 + x²). Separating: dy/(1 + y²) = dx/(1 + x²), so tan⁻¹ y = tan⁻¹ x + C.
Example: growth. The rate of growth of a bacterial population is proportional to its size: dP/dt = kP. Then dP/P = k dt, log P = kt + C, so P = P₀ e^(kt), where P₀ is the population at t = 0. If the population doubles in 3 hours, k = (log 2)/3.
Method 2: homogeneous equations
A function F(x, y) is homogeneous of degree zero if F(λx, λy) = F(x, y) for all λ. For example, (x² + y²)/(xy) is homogeneous of degree zero; x² + y is not homogeneous.
An equation dy/dx = F(x, y) with F homogeneous of degree zero is solved by the substitution y = vx, so that dy/dx = v + x dv/dx. This turns it into a variables separable equation in v and x.
Example. dy/dx = (x + y)/x. Put y = vx: v + x dv/dx = 1 + v, so x dv/dx = 1, dv = dx/x, v = log |x| + C. Since v = y/x, y = x log |x| + Cx.
Method 3: linear first-order equations
An equation of the form
dy/dx + P y = Q, where P and Q are functions of x only (or constants),
is a linear differential equation. Multiply both sides by the integrating factor
I.F. = e^(∫P dx)
The left side becomes the derivative of y × I.F., and the solution is
y × (I.F.) = ∫ Q × (I.F.) dx + C
Example. dy/dx + y = eˣ. Here P = 1, Q = eˣ. I.F. = eˣ. So y eˣ = ∫ e^(2x) dx = e^(2x)/2 + C, giving y = eˣ/2 + C e⁻ˣ.
Example. x (dy/dx) + 2y = x², for x > 0. Divide by x: dy/dx + (2/x) y = x. I.F. = e^(2 log x) = x². So y x² = ∫ x³ dx = x⁴/4 + C, giving y = x²/4 + C/x².
Some equations are linear in x instead: dx/dy + P₁ x = Q₁, with P₁ and Q₁ functions of y. Then I.F. = e^(∫P₁ dy) and x × I.F. = ∫ Q₁ × I.F. dy + C.
Finding a particular solution
Solve generally, then use the given condition to find C.
Example. dy/dx = 2x, with y = 3 when x = 1. General solution: y = x² + C. Putting x = 1, y = 3: C = 2. Particular solution: y = x² + 2.
Summary
Order = highest derivative; degree = its power (when the equation is polynomial in derivatives).
The general solution has as many arbitrary constants as the order.
Separable: separate and integrate. Homogeneous: substitute y = vx.
Linear dy/dx + Py = Q: multiply by e^(∫P dx); y × I.F. = ∫ Q × I.F. dx + C.