First-Order Differential Equations

What a first-order ODE is

An ordinary differential equation is an equation involving an unknown function and its derivatives. "First-order" means only the first derivative appears. The most general form is

y' = f(x, y).

You can read this geometrically: at every point (x, y), the equation prescribes a slope f(x, y). Drawing those slopes as little dashes is a slope field — and a solution is any curve whose tangent always matches the local dash. An initial value problem (IVP) pins down a specific solution by fixing one point: y(x₀) = y₀.

Most physical processes that involve "rate of change" produce first-order ODEs: radioactive decay (dN/dt = −λN), Newton's law of cooling (dT/dt = −k(T − T_room)), simple population growth (dP/dt = rP), and RC circuits all fit this template.

Type 1: Separable

The right side factors as a product y' = g(x) h(y). Move all y to one side and all x to the other:

dy / h(y) = g(x) dx,

integrate both sides, and solve for y. Worked example — exponential growth:

dy/dx = y → dy/y = dx → ln|y| = x + C → y = Ce^x.

An initial condition y(0) = y₀ determines C = y₀, giving y = y₀ e^x. Every "doubling time" or "half-life" computation reduces to this.

Type 2: Linear

A first-order linear ODE has the standard form y' + p(x) y = q(x). It's "linear in y and y'" with no y² or y · y' terms. The trick is the integrating factor:

μ(x) = exp(∫ p(x) dx).

Multiplying through by μ turns the left side into a single derivative:

(μy)' = μ q(x),

so y = (1/μ) [ ∫ μ(x) q(x) dx + C ]. Worked example: y' + 2y = e^{−x}. Here p = 2, μ = e^{2x}. Then (e^{2x} y)' = e^{2x} · e^{−x} = e^x, so e^{2x} y = e^x + C and y = e^{−x} + C e^{−2x}.

Type 3: Exact

An equation in differential form M(x,y) dx + N(x,y) dy = 0 is exact when there exists a potential function F(x,y) with ∂F/∂x = M and ∂F/∂y = N. The exactness test:

∂M/∂y = ∂N/∂x.

If true, integrate M with respect to x to get F(x,y) = ∫ M dx + g(y), then differentiate with respect to y and match to N to find g(y). The solutions are level curves F(x, y) = C. If the test fails, you can sometimes find an integrating factor μ(x,y) that makes it exact.

Type 4: Bernoulli

A Bernoulli equation has the form y' + p(x) y = q(x) y^n with n ≠ 0, 1 (those reduce to linear or separable). The substitution u = y^{1−n} turns it into a linear ODE in u:

u' + (1 − n) p(x) u = (1 − n) q(x).

Solve with the integrating-factor method, then back-substitute y = u^{1/(1−n)}. The classic example is the logistic equation y' = ay − by², which is Bernoulli with n = 2.

Type 5: Riccati

A Riccati equation is y' = p(x) + q(x) y + r(x) y² — a first-order ODE with a quadratic right side. There's no general closed-form solution, but if you happen to know one particular solution y₁(x), the substitution y = y₁ + 1/u turns it into a linear equation in u. Riccati equations show up in optimal control (algebraic Riccati equation) and in the WKB approximation of quantum mechanics.

Five types compared

Type	Form	Recipe	Example
Separable	y' = g(x) h(y)	Separate, integrate both sides	y' = y → y = Ce^x
Linear	y' + p(x) y = q(x)	μ = e^{∫p dx}, multiply, integrate	y' + y = x → y = (x−1) + Ce^{−x}
Exact	M dx + N dy = 0 with M_y = N_x	Find F with F_x = M, F_y = N; F = C	(2xy + y²) dx + (x² + 2xy) dy = 0
Bernoulli	y' + p y = q y^n	u = y^{1−n}; reduces to linear	y' = y − y² (logistic)
Riccati	y' = p + qy + ry²	If y₁ known, y = y₁ + 1/u → linear	y' = y² − x²
Homogeneous (degree)	y' = f(y/x)	v = y/x reduces to separable	y' = (x+y)/(x−y)

Worked application: Newton's law of cooling

A coffee at temperature T(t) sits in a room at T_r. The cooling rate is proportional to the difference:

dT/dt = −k (T − T_r).

This is separable. Let u = T − T_r; then du/dt = −ku, so u = u₀ e^{−kt}. In original variables,

T(t) = T_r + (T₀ − T_r) e^{−kt}.

The temperature approaches T_r exponentially, with time constant 1/k. The half-life of the temperature gap is ln(2)/k. The same equation governs RC circuit decay (with T replaced by voltage) and radioactive cooldown — first-order linear behavior is everywhere.

Worked application: logistic growth

Population growth with a carrying capacity K obeys

dP/dt = r P (1 − P/K).

This is Bernoulli with n = 2, but it's also separable. After partial fractions and integration,

P(t) = K / (1 + A e^{−rt}),

where A = (K − P₀)/P₀. Early on (P ≪ K) it grows exponentially; late on, it saturates at K. The same equation models autocatalytic reactions, viral spread, and adoption curves.

Existence and uniqueness

The Picard–Lindelöf theorem guarantees that an IVP y' = f(x, y), y(x₀) = y₀ has a unique solution on some interval around x₀ provided f and ∂f/∂y are continuous near (x₀, y₀). Without continuity of ∂f/∂y, multiple solutions can pass through the same initial point — the classic example is y' = √y, which has both y = (x/2)² and y = 0 starting from (0, 0).

Common mistakes

• Sign errors when separating variables. When integrating dy/y, you get ln|y|, not ln(y). The absolute value matters whenever y can change sign.
• Forgetting the constant of integration. Every first-order ODE has a one-parameter family of solutions. Drop the +C and you've lost all but one curve.
• Missing equilibrium solutions. Dividing by h(y) in a separable ODE assumes h(y) ≠ 0. Roots of h(y) = 0 give constant solutions you must check separately.
• Wrong integrating factor. The standard form for the linear method is y' + p(x) y = q(x) — coefficient of y' must be 1 before computing μ. Otherwise divide first.
• Skipping the exactness test. Trying to find a potential F before checking ∂M/∂y = ∂N/∂x wastes time on equations that aren't exact. The two-line check costs nothing.
• Bernoulli with wrong substitution. Use u = y^{1−n}, not u = y^n. Getting the sign of (1 − n) wrong flips the sign on every coefficient in the resulting linear equation.