First-Order ODEs

⚡ Key Takeaways

First-order ODEs are differential equations involving only the first derivative of an unknown function: $y' = f(x,y)$ . Key result: a vast range of such equations can be solved exactly by recognizing their type (separable, linear, exact, etc.), and the Picard–Lindelöf Theorem guarantees a unique solution near any point where $f$ is reasonably smooth. Where it applies: population growth, radioactive decay, RC circuits, Newtonian motion with resistance, chemical kinetics, mixing problems. Who/when: Newton, Leibniz, the Bernoullis, Euler, and Clairaut founded the subject between roughly 1670 and 1750.

💡 The Core Insight

A first-order ODE assigns to every point $(x,y)$ in the plane a slope $f(x,y)$ . Solving the equation means finding curves that are everywhere tangent to this prescribed direction field. Every solution method — separation, integrating factors, substitutions — is a trick for converting this geometric tangency condition into an antiderivative we can actually compute.

DefinitionFirst-Order ODE in Normal Form

A first-order ordinary differential equation in normal (explicit) form is

1

\quad \frac{dy}{dx} = f(x,y),

Derivation hint: this is the definition; $f$ is any given function of two variables, $x$ the independent variable, $y=y(x)$ the unknown function, $dy/dx$ its derivative. A function $y=\varphi(x)$ defined on an interval $I$ is a solution if it is differentiable on $I$ and $\varphi'(x)=f(x,\varphi(x))$ for all $x\in I$ . An initial value problem (IVP) pairs equation [1] with a condition

2

\quad y(x_0) = y_0,

Derivation hint: this pins down one curve from the family by requiring the solution to pass through the prescribed point $(x_0,y_0)$ . The general solution is a family $y=\varphi(x,C)$ depending on one arbitrary constant $C$ ; a particular solution fixes $C$ .

DefinitionThe Four Canonical Solvable Types

Separable: $\dfrac{dy}{dx}=g(x)h(y)$ .

Linear: $\dfrac{dy}{dx}+p(x)\,y = q(x)$ .

Exact: $M(x,y)\,dx + N(x,y)\,dy = 0$ with $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$ .

Homogeneous (in the scaling sense): $\dfrac{dy}{dx}=F\!\left(\dfrac{y}{x}\right)$ , solved by the substitution $v=y/x$ .

TheoremPicard–Lindelöf Existence and Uniqueness

Let $f(x,y)$ be continuous on a rectangle $R=\{|x-x_0|\le a,\ |y-y_0|\le b\}$ and satisfy a Lipschitz condition in $y$ :

3

\quad |f(x,y_1)-f(x,y_2)| \le L\,|y_1-y_2| \quad \text{for all } (x,y_1),(x,y_2)\in R.

Derivation hint: the Lipschitz constant $L$ bounds how fast $f$ changes in the $y$ -direction; it follows automatically if $\partial f/\partial y$ is continuous and bounded on $R$ . Then the IVP [1]–[2] has a unique solution on the interval $|x-x_0|\le h$ , where $h=\min(a,\,b/M)$ and $M=\max_R|f|$ .

TheoremLinear Equations: The Integrating Factor Solution

The general linear equation $y'+p(x)y=q(x)$ has the explicit solution

4

\quad y(x) = \frac{1}{\mu(x)}\left[\int \mu(x)\,q(x)\,dx + C\right], \qquad \mu(x) = e^{\int p(x)\,dx}.

Derivation hint: choose $\mu$ so the left side becomes the derivative of a product $(\mu y)'$ ; see the proof below.

ProofIntegrating Factor for Linear Equations

We solve $y'+p(x)y=q(x)$ by forcing the left side to be a perfect derivative.

Step 1. Multiply the entire equation by an as-yet-unknown positive function $\mu(x)$ :

5

\quad \mu y' + \mu p\, y = \mu q.

Derivation hint: multiply both sides of $y'+py=q$ by $\mu$ .

Step 2. We want the left side to equal $\dfrac{d}{dx}(\mu y) = \mu y' + \mu' y$ . Comparing with [5], this requires

6

\quad \mu'(x) = p(x)\,\mu(x).

Derivation hint: match the coefficient of $y$ in $\mu p\,y$ with the coefficient in $\mu' y$ .

Step 3. Equation [6] is itself separable: $d\mu/\mu = p\,dx$ , so integrating gives

7

\quad \mu(x) = e^{\int p(x)\,dx}.

Derivation hint: integrate $\ln\mu = \int p\,dx$ and exponentiate; the constant of integration may be set to 1 since any nonzero multiple of $\mu$ works.

Step 4. With this $\mu$ , equation [5] becomes $(\mu y)' = \mu q$ . Integrate both sides:

8

\quad \mu(x)\,y(x) = \int \mu(x)\,q(x)\,dx + C.

Derivation hint: antidifferentiate $(\mu y)' = \mu q$ with respect to $x$ , adding the constant $C$ .

Step 5. Divide by $\mu$ to obtain [4]. Uniqueness of the constant $C$ follows because two antiderivatives differ by a constant. $\blacksquare$

∎

ProofSeparable Equations

Given $dy/dx = g(x)h(y)$ with $h(y)\ne 0$ , divide and integrate:

9

\quad \int \frac{dy}{h(y)} = \int g(x)\,dx + C.

Derivation hint: rewrite as $\frac{1}{h(y)}\,dy = g(x)\,dx$ and integrate each side independently; the single constant $C$ absorbs both integration constants. Each root of $h(y)=0$ gives an additional equilibrium (constant) solution that this division may have discarded. $\blacksquare$

∎

ProofExact Equations

Suppose $M\,dx + N\,dy = 0$ with $M_y = N_x$ on a simply connected region. Then there exists a potential function $\Psi(x,y)$ with $\Psi_x = M$ and $\Psi_y = N$ . Along a solution curve $d\Psi = M\,dx + N\,dy = 0$ , so the solution is the implicit relation

10

\quad \Psi(x,y) = C, \qquad \Psi(x,y) = \int M\,dx + \int\!\left(N - \frac{\partial}{\partial y}\!\int M\,dx\right)dy.

Derivation hint: integrate $\Psi_x=M$ in $x$ (treating $y$ constant, picking up an unknown function $g(y)$ ), then differentiate the result in $y$ and match against $\Psi_y=N$ to determine $g'(y)$ . The condition $M_y=N_x$ is exactly the Clairaut equality of mixed partials $\Psi_{xy}=\Psi_{yx}$ , and is what makes $\Psi$ exist. $\blacksquare$

∎

mathematician · Leonhard Euler | 1707–1783

Euler systematized the entire subject. He introduced the integrating factor as a general device (around 1734–1740), classified exact equations and derived the exactness criterion $M_y=N_x$ , invented the numerical Euler method that still bears his name, and treated the substitution methods for homogeneous and Bernoulli-type equations. His Institutionum calculi integralis (1768–1770) collected these results into the first coherent textbook treatment, turning a bag of clever tricks into a recognizable theory.

🧠 Intuition

Picture the right side $f(x,y)$ as a recipe that, at every point of the plane, hands you a little arrow telling you which way to head. A solution is a road that always points along the local arrows — it can never cross them transversally. This is why uniqueness fails exactly where arrows pile up ambiguously: if two roads could leave the same point, the arrow there would be undefined. For the linear equation, the integrating factor $\mu=e^{\int p}$ is best understood as a change of unknown variable. Setting $w=\mu y$ , the cluttered equation collapses to $w'=\mu q$ , which is just "integrate once." The factor $\mu$ measures the cumulative exponential drift the system would undergo if undriven ( $q=0$ ), and dividing by it "undoes" that drift so the remaining driving term integrates cleanly. For separable equations, the insight is that $x$ and $y$ have been completely decoupled: each lives on its own side of the equation, so the problem splits into two independent one-variable integrals.

Figure

Direction field of

y'=x-y

with one solution curve (gold) threading the arrows

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ExampleLinear: an RC circuit driven by a battery

Given: A capacitor charges through a resistor. Kirchhoff's law gives $R\dfrac{dq}{dt}+\dfrac{q}{C}=V_0$ , with $R=2\ \Omega$ , $C=0.5\ \text{F}$ , $V_0=10\ \text{V}$ , and initial charge $q(0)=0\ \text{C}$ .

Step 1. Put in standard linear form. Divide by $R=2$ :

11

\quad \frac{dq}{dt} + \frac{1}{RC}\,q = \frac{V_0}{R} = \frac{1}{1}\,q\text{-coeff }= 1\cdot q,\quad \text{i.e. } \frac{dq}{dt}+1\cdot q = 5.

Derivation hint: $1/(RC) = 1/(2\cdot 0.5)=1$ and $V_0/R = 10/2 = 5$ .

Step 2. Integrating factor: $p(t)=1$ , so $\mu = e^{\int 1\,dt}=e^{t}$ .

Step 3. Apply [4]:

12

\quad q(t) = e^{-t}\left[\int 5\,e^{t}\,dt + C\right] = e^{-t}\big[5e^{t}+C\big] = 5 + Ce^{-t}.

Derivation hint: $\int 5e^{t}\,dt = 5e^{t}$ ; then multiply through by $e^{-t}$ .

Step 4. Apply $q(0)=0$ : $0 = 5 + C \Rightarrow C=-5$ .

\boxed{\,q(t) = 5\big(1-e^{-t}\big)\ \text{C}\,}

The charge rises to a steady-state $5\ \text{C}$ (i.e. $CV_0$ ) with time constant $\tau = RC = 1\ \text{s}$ .

ExampleSeparable: Newton's law of cooling

Given: A cup of coffee at $90^\circ$ C cools in a $20^\circ$ C room. Newton's law: $\dfrac{dT}{dt}=-k(T-20)$ with measured $k=0.1\ \text{min}^{-1}$ . Find $T(t)$ and the time to reach $50^\circ$ C.

Step 1. Separate variables:

13

\quad \frac{dT}{T-20} = -k\,dt.

Derivation hint: divide both sides by $(T-20)$ and multiply by $dt$ .

Step 2. Integrate [9]: $\ln|T-20| = -kt + C_1$ , so $T-20 = Ce^{-kt}$ . Derivation hint: exponentiate; $C=\pm e^{C_1}$ absorbs the constant and sign.

Step 3. Initial condition $T(0)=90$ : $90-20 = C \Rightarrow C=70$ . Thus

14

\quad T(t) = 20 + 70\,e^{-0.1\,t}\ \ (^\circ\text{C}).

Derivation hint: substitute $C=70$ , $k=0.1$ into $T=20+Ce^{-kt}$ .

Step 4. Solve $T=50$ : $30 = 70e^{-0.1t}\Rightarrow e^{-0.1t}=3/7 \Rightarrow t = -10\ln(3/7)$ .

\boxed{\,t = 10\ln(7/3) \approx 8.47\ \text{min}\,}

⚠️ Common mistake

"The general solution captures every solution." It usually does not. When dividing by $h(y)$ in a separable equation you may lose equilibrium solutions where $h(y)=0$ . For $y'=y$ the general solution $y=Ce^{x}$ does include $y\equiv0$ (set $C=0$ ), but for $y'=y^2$ the constant $y\equiv0$ is a genuine solution not obtained from $y=-1/(x+C)$ for any finite $C$ . Always check the roots of $h$ separately.

"A solution exists for all $x$ ." False in general. Even with perfectly smooth $f$ , solutions can blow up in finite time (see the counterexample). Picard–Lindelöf only guarantees a local interval.

" $+C$ is decorative." The constant is the solution's identity card — it selects which member of the one-parameter family you mean. Dropping it loses all but one solution.

CounterexampleFinite-time blow-up and non-uniqueness

Blow-up: The IVP $y'=y^2$ , $y(0)=1$ has $f=y^2$ smooth everywhere, yet its solution is

15

\quad y(x)=\frac{1}{1-x},

Derivation hint: separable; $\int y^{-2}dy = \int dx$ gives $-1/y = x - 1$ . which escapes to $+\infty$ as $x\to 1^-$ . No solution exists for $x\ge 1$ — global existence fails despite smoothness.

Non-uniqueness: The IVP $y'=y^{1/3}$ , $y(0)=0$ violates the Lipschitz condition at $y=0$ (the derivative $\tfrac13 y^{-2/3}$ blows up there). It admits infinitely many solutions, including $y\equiv 0$ and $y=\big(\tfrac{2}{3}x\big)^{3/2}$ , plus glued combinations. Continuity of $f$ alone (Peano's theorem) buys existence but not uniqueness.

connection

First-order ODEs are the foundation on which the rest of differential equations is built. Any $n$ -th order equation $y^{(n)}=g(x,y,\dots,y^{(n-1)})$ reduces to a first-order system by setting $u_1=y,\,u_2=y',\dots$ , so the theory here is the atom of the whole subject — see Higher-Order Linear ODEs and Systems of First-Order ODEs. The integrating-factor idea reappears as the variation of parameters technique. The exactness criterion $M_y=N_x$ is the two-dimensional shadow of the curl-free / conservative-field condition in Vector Calculus and of closed-versus-exact differential forms. The direction-field viewpoint and equilibrium analysis lead directly into Autonomous Equations and Phase Lines and Dynamical Systems. When no closed form exists, one turns to Numerical Methods for ODEs (Euler, Runge–Kutta), whose simplest member Euler invented for exactly this class.

🔬 Where this appears

⏳ History

- 1671–1676 — Newton: In Method of Fluxions, Newton solves "fluxional equations" by power series, the first systematic attack on first-order ODEs.

- 1684–1693 — Leibniz: Develops the

dy/dx

notation, names the "differential equation," and discovers separation of variables and the method for homogeneous equations.

- 1694 — Leibniz & Johann Bernoulli: The integrating factor technique for linear equations is recognized.

- 1696 — The Bernoullis: Johann poses the brachistochrone; Jakob's earlier work leads to the Bernoulli equation

y'+p(x)y=q(x)y^n

, linearized by substitution.

- 1724 — Riccati: Jacopo Riccati studies

y'=q_0+q_1y+q_2y^2

, the simplest equation not generally solvable in elementary terms.

- 1734 — Clairaut: Introduces the exactness criterion and singular solutions (the Clairaut equation

y=xy'+f(y')

).

- 1768–1770 — Euler: Institutionum calculi integralis unifies integrating factors, exact equations, substitutions, and the Euler numerical method.

- 1820s–1830s — Cauchy: Founds the rigorous existence theory via the method of successive approximation.

- 1890–1894 — Picard & Lindelöf: Sharpen Cauchy's result into the modern existence–uniqueness theorem using the Lipschitz condition and the contraction principle.

open-problem

The classification into solvable types is deceptively incomplete. Most first-order ODEs have no closed-form solution in elementary functions — this is not a gap in our cleverness but a structural fact, made precise by differential Galois theory (Liouville, Risch). The Riccati equation $y'=q_0+q_1y+q_2y^2$ is the boundary case: it is solvable in closed form only when one particular solution is already known, and otherwise links to second-order linear equations and special functions. Singular solutions — envelopes of the general family, as in the Clairaut equation — are not captured by the constant $C$ at all, and a complete catalog of when they arise remains subtle. Finally, the gap between continuity (Peano: existence) and Lipschitz (Picard: uniqueness) hides a rich landscape; characterizing the exact regularity needed for uniqueness, and understanding the set of solutions when it fails, connects to active research in PDE theory and the analysis of ill-posed problems.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also