First-Order Differential Equations

A differential equation relates a function to its derivatives. Its order is the highest derivative that appears; its degree is the power of that highest derivative once the equation is free of radicals and fractions in the derivatives. The general solution of an n-th order equation has n arbitrary constants, fixed by initial or boundary conditions.

Variable separable

dy/dx = f(x)g(y) ⇒ ∫ dy/g(y) = ∫ f(x) dx.

Example. dy/dx = ky gives ln y = kx + c, so y = Ce^kx: exponential growth (k > 0) or decay (k < 0). Newton's law of cooling dT/dt = −k(T − T_a) gives T = T_a + (T₀ − T_a)e^−kt.

Homogeneous equations

dy/dx = f(y/x): substitute y = vx, so dy/dx = v + x dv/dx, which separates.

Linear first-order equations

dy/dx + P(x)y = Q(x). Multiply by the integrating factor IF = e^{∫P dx}:

y · IF = ∫ Q · IF dx + C

Example. dy/dx + y = x. IF = eˣ, so yeˣ = ∫ xeˣ dx = eˣ(x − 1) + C, and y = x − 1 + Ce^−x. With y(0) = 0, C = 1.

Exact equations

M dx + N dy = 0 is exact if ∂M/∂y = ∂N/∂x. Then there is a function f with ∂f/∂x = M and ∂f/∂y = N, and the solution is f(x, y) = C. If it is not exact, an integrating factor may make it so: if (∂M/∂y − ∂N/∂x)/N depends on x only, IF = e^{∫ that dx}.

Bernoulli equations

dy/dx + P(x)y = Q(x)yⁿ (n ≠ 0, 1). Substituting z = y¹⁻ⁿ turns it into a linear equation in z.

Worked example

Solve dy/dx + 2y = 4 with y(0) = 0 exactly (y = 2(1 − e^−2x)), and compare with a numerical solution. Check whether (2xy + 3) dx + (x² + 4y) dy = 0 is exact.

import math

# Numerical check with many small Euler steps
y, x, h = 0.0, 0.0, 1e-5
while x < 1 - 1e-12:
    y += h * (4 - 2 * y)
    x += h
print(f"numerical y(1) = {y:.5f}, exact 2(1 - e^-2) = {2 * (1 - math.exp(-2)):.5f}")

# Exactness: M = 2xy + 3, N = x^2 + 4y -> dM/dy = 2x, dN/dx = 2x
dMdy = lambda x, y: 2 * x
dNdx = lambda x, y: 2 * x
print("exact:", all(dMdy(a, b) == dNdx(a, b) for a in range(-3, 4) for b in range(-3, 4)))
print("solution: x^2 y + 3x + 2y^2 = C")

# Cooling: body at 90 C in a 30 C room, 60 C after 10 min. Temperature after 20 min?
k = math.log((90 - 30) / (60 - 30)) / 10
print(f"k = {k:.5f} per min, T(20) = {30 + 60 * math.exp(-20 * k):.2f} C")

numerical y(1) = 1.72934, exact 2(1 - e^-2) = 1.72933
exact: True
solution: x^2 y + 3x + 2y^2 = C
k = 0.06931 per min, T(20) = 45.00 C

Practice questions