First-Order Differential Equations

Solving first-order differential equations using separable variables, integrating factors, and linear methods.

What it is

A first-order ordinary differential equation relates an independent variable $x$, an unknown function $y(x)$, and its first derivative $y'(x)$: $F(x, y, y') = 0 \quad \implies \quad \frac{dy}{dx} = f(x, y)$ Depending on the algebraic structure of $f(x, y)$, they fall into several analytically solvable categories:

1. Separable: The derivative factors into independent single-variable functions: $\frac{dy}{dx} = g(x) h(y) \implies \int \frac{dy}{h(y)} = \int g(x)\,dx + C$
2. First-order linear: Both $y$ and $y'$ appear linearly: $\frac{dy}{dx} + P(x) y = Q(x)$ Solved by multiplying through by the canonical integrating factor $\mu(x) = \exp\left(\int P(x)\,dx\right)$.
3. Exact equations: The differential form $M(x, y)\,dx + N(x, y)\,dy = 0$ is the total exterior derivative of a scalar potential $\Phi(x, y)$: $d\Phi = \frac{\partial \Phi}{\partial x}\,dx + \frac{\partial \Phi}{\partial y}\,dy = 0 \iff \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ The implicit solution family consists of level contours $\Phi(x, y) = C$.
4. Homogeneous of degree zero: Satisfies $f(\lambda x, \lambda y) = f(x, y)$, reduced to separable form via substitution $u = y/x$.
5. Bernoulli equations: Nonlinear equations $y' + P(x) y = Q(x) y^n$ ($n \neq 0, 1$), transformed into linear ODEs via substitution $u = y^{1-n}$.

Why it appears in physics

Any relaxation, charging, or dissipation process where the rate of change is proportional to current state level is governed by a first-order ODE:

• Capacitor charging/discharging (RC circuit): By Kirchhoff's loop rule, $R \frac{dq}{dt} + \frac{1}{C} q = V(t)$.
• Inductor transient response (RL circuit): $L \frac{di}{dt} + R i = V(t)$.
• Newton's law of convective cooling: $\frac{dT}{dt} = -k(T - T_{\text{env}})$.
• Terminal velocity with Stokes drag: $m \frac{dv}{dt} = m g - \gamma v$.
• Radiation attenuation (Beer-Lambert law): Light intensity drops across optical thickness: $\frac{dI}{dx} = -\mu(\lambda) I$.

Visual or geometric intuition

A first-order ODE defines a tangent vector field across every point in the $(x, y)$ plane: $\mathbf{v}(x, y) = \begin{pmatrix} 1 \\ f(x, y) \end{pmatrix}$ The Picard-Lindelöf theorem guarantees that away from singular branch points, exactly one integral curve passes through each point. Integral curves cannot intersect or cross in the plane, behaving exactly like steady streamlines in a 2D incompressible laminar fluid flow.

Minimal symbolic language

• Pfaffian differential form: $\omega = M(x, y)\,dx + N(x, y)\,dy = 0$
• Integrability condition (Schwarz symmetry / exactness criterion): $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
• Euler integrating factor: If not exact, a multiplier $\mu(x, y)$ such that $(\mu M)\,dx + (\mu N)\,dy = 0$ is exact satisfies: $M \frac{\partial \mu}{\partial y} - N \frac{\partial \mu}{\partial x} = \mu \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)$
• Closed-form solution for first-order linear ODE: $y(x) = \frac{1}{\mu(x)} \left[\int \mu(x) Q(x)\,dx + C\right], \quad \mu(x) = e^{\int P(x)\,dx}$

What the result means

1. Transient vs. steady-state decomposition: In stable systems with $P(x) = a > 0$, the solution splits into an exponentially vanishing transient ($C e^{-a t} \to 0$) and a forced steady-state response driven by $Q(t)$.
2. Characteristic relaxation time $\tau$: For first-order dynamics, $\tau = 1/P$ sets the natural relaxation timescale: at $t = \tau$, $63.2\%$ of the step response is accomplished; at $t = 5\tau$, steady state is practically reached ($99.3\%$).
3. Smooth foliation: The parameter $C$ sweeps a smooth 1-parameter family of non-intersecting trajectories across state space.

When not to use it

• Inertial oscillatory systems: Autonomous 1D first-order ODEs are strictly monotonic; they cannot exhibit free oscillations or closed orbits without an explicit time-dependent periodic driver. True harmonic resonance requires second order.
• Ignoring equilibrium roots: When dividing by $h(y)$ during separation of variables, omitting roots $h(y^*) = 0$ discards constant stationary states.
• Treating Bernoulli equations as linear: Forgetting to substitute $u = y^{1-n}$ leads to invalid superposition.

Operational summary

1. Normalize to standard form: $\frac{dy}{dx} = f(x, y)$ or $M\,dx + N\,dy = 0$.
2. Check separability: Can you factor $f(x, y) = g(x) h(y)$? If yes, separate and integrate.
3. Check linearity: Does it fit $y' + P(x) y = Q(x)$? If yes, compute $\mu(x) = e^{\int P\,dx}$ and integrate.
4. Check exactness: Does $\partial_y M = \partial_x N$? If yes, integrate $M\,dx$ and determine $h(y)$.
5. If inexact, test for integrating factors depending on $x$ only: $\frac{\partial_y M - \partial_x N}{N} = f(x)$.
6. For Bernoulli forms $y' + P y = Q y^n$, linearize via $u = y^{1-n}$.

Procedure: Systematic Solution of First-Order ODEs

Before calculating

1. Isolate the derivative: Rewrite in standard form $\frac{dy}{dx} = f(x, y)$ or symmetric differential form $M(x, y)\,dx + N(x, y)\,dy = 0$.
2. Confirm degree: Ensure derivatives are not raised to fractional or higher powers like $(y')^2$.
3. Inspect denominator roots: If dividing by any function of $y$, record zero roots beforehand (potential stationary equilibrium states).

Step 1: Identify the mathematical object

Classify the ODE following the decision hierarchy:

• Separable? $\frac{dy}{dx} = g(x) h(y)$.
• Linear? $\frac{dy}{dx} + P(x) y = Q(x)$.
• Exact? $M\,dx + N\,dy = 0$ with $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.
• Homogeneous? $M(\lambda x, \lambda y) = \lambda^k M$ and $N(\lambda x, \lambda y) = \lambda^k N$.
• Bernoulli? $\frac{dy}{dx} + P(x) y = Q(x) y^n$.

Step 2: Choose the operation

• If Separable: Rearrange: $\frac{dy}{h(y)} = g(x)\,dx$ and integrate: $\int \frac{dy}{h(y)} = \int g(x)\,dx + C$.
• If Linear:
  1. Compute integrating factor: $\mu(x) = \exp\left(\int P(x)\,dx\right)$. (Omit $+C$ here).
  2. Multiply the equation by $\mu(x)$, combining the left side into $\frac{d}{dx}[\mu(x) y]$.
  3. Integrate both sides: $\mu(x) y(x) = \int \mu(x) Q(x)\,dx + C$.
  4. Isolate $y(x) = \frac{1}{\mu(x)}\int \mu(x) Q(x)\,dx + \frac{C}{\mu(x)}$.
• If Exact:
  1. Integrate $M$ with respect to $x$: $\Phi(x, y) = \int M(x, y)\,dx + g(y)$.
  2. Differentiate with respect to $y$ and equate to $N$: $\frac{\partial \Phi}{\partial y} = N(x, y)$.
  3. Integrate $g'(y)$ to find $g(y)$.
  4. Write the implicit solution: $\Phi(x, y) = C$.
• If Bernoulli ($y' + P y = Q y^n$):
  1. Divide by $y^n$: $y^{-n} y' + P y^{1-n} = Q$.
  2. Substitute $u = y^{1-n} \implies u' = (1-n) y^{-n} y'$.
  3. Solve the resulting linear ODE: $\frac{1}{1-n} u' + P(x) u = Q(x)$.

Step 3: Conserve units and signs

• In $\mu(x) = e^{\int P(x) dx}$, ensure the exponent is dimensionless. If $x$ is in seconds, $P(x)$ must be in $\text{s}^{-1}$.
• In electrical circuits, preserve correct polarity across resistors and power sources.

Step 4: Calculate

1. Evaluate primitives using standard calculus rules.
2. Embed the integration constant $+C$ immediately upon performing indefinite quadratures.

Step 5: Interpret the result

1. Apply the initial condition $y(x_0) = y_0$ to fix $C$.
2. Analyze asymptotic stability ($t \to \infty$): does the transient vanish while tracking the steady state?
3. Confirm dimensional consistency across all terms.

Self-check checklist

• Did I test exactness by calculating both $\partial_y M$ and $\partial_x N$?
• Did I insert $+C$ prior to isolating $y$?
• Did I check whether $h(y) = 0$ yielded singular solutions?
• Is the integrating factor exponent dimensionless?
• Does substituting the solution back into the ODE yield an identity?

Worked Examples: First-Order Equations

Step-by-step guided example

Problem

A series electrical circuit consists of a resistor $R = 50\,\Omega$, an inductor $L = 2.5\,\text{H}$, and a DC voltage source $V_0 = 100\,\text{V}$. At initial instant $t = 0$, the switch is closed with zero initial current: $i(0) = 0\,\text{A}$.

1. Formulate the differential equation for current $i(t)$.
2. Solve the ODE using the integrating factor technique.
3. Determine the characteristic time constant $\tau$, steady-state current $i_\infty$, and the time required to reach $90\%$ of full steady current.

Detailed solution

Step 1: Governing ODE formulation Applying Kirchhoff's voltage law around the loop: $L \frac{di}{dt} + R i = V_0 \implies \frac{di}{dt} + \frac{R}{L} i = \frac{V_0}{L}$ Substituting numerical values: $\frac{di}{dt} + \frac{50}{2.5} i = \frac{100}{2.5} \implies \frac{di}{dt} + 20 i = 40$ This is a first-order non-homogeneous linear ODE with $P(t) = 20\,\text{s}^{-1}$ and $Q(t) = 40\,\text{A}/\text{s}$.

Step 2: Integrating factor solution The integrating factor is: $\mu(t) = \exp\left(\int 20\,dt\right) = e^{20 t}$ Multiplying through: $e^{20 t} \frac{di}{dt} + 20 e^{20 t} i = 40 e^{20 t} \implies \frac{d}{dt}\left(i(t) e^{20 t}\right) = 40 e^{20 t}$ Integrating both sides: $i(t) e^{20 t} = \int 40 e^{20 t}\,dt = \frac{40}{20} e^{20 t} + C = 2 e^{20 t} + C$ Isolating $i(t)$: $i(t) = 2 + C e^{-20 t}$

Step 3: Imposing the initial condition At $t = 0$, $i(0) = 0$: $0 = 2 + C e^0 = 2 + C \implies C = -2\,\text{A}$ The particular solution is: $i(t) = 2\left(1 - e^{-20 t}\right)\,\text{A}$

Step 4: Physical parameter evaluation

• Steady-state current: As $t \to \infty$, $e^{-20 t} \to 0$: $i_\infty = 2\,\text{A} = \frac{V_0}{R} = \frac{100}{50} = 2\,\text{A}$
• Inductive time constant $\tau$: $\tau = \frac{L}{R} = \frac{2.5}{50} = 0.05\,\text{s} = 50\,\text{ms}$ (The rate constant matches $1/\tau = 20\,\text{s}^{-1}$).
• Time to reach $90\%$ of steady state ($i = 1.8\,\text{A}$): $1.8 = 2(1 - e^{-20 t}) \implies 0.1 = e^{-20 t} \implies t = \frac{\ln 10}{20} \approx 0.1151\,\text{s} = 115.1\,\text{ms} \approx 2.3\tau$

Inverse example

Problem

In a colloidal sedimentation experiment, the depth profile of particle concentration $C(x)$ ($x \ge 0$) follows: $C(x) = \frac{C_0}{1 + k x C_0}$ where $C_0$ is the surface concentration at $x = 0$ and $k > 0$ is an agglomeration coefficient. Demonstrate that this distribution satisfies a non-linear first-order ODE (Bernoulli / separable type) and derive that governing differential equation.

Detailed solution

Step 1: Differentiating the profile Compute $\frac{dC}{dx}$: $\frac{dC}{dx} = \frac{d}{dx}\left[C_0(1 + k x C_0)^{-1}\right] = -k C_0^2 (1 + k x C_0)^{-2}$

Step 2: Expressing in terms of state variable $C$ Observe that: $C(x) = \frac{C_0}{1 + k x C_0} \implies C^2(x) = \frac{C_0^2}{(1 + k x C_0)^2}$ Substitute into the derivative expression: $\frac{dC}{dx} = -k \left(\frac{C_0^2}{(1 + k x C_0)^2}\right) = -k C^2$

Step 3: Differential equation conclusion The governing differential equation is: $\frac{dC}{dx} + k C^2 = 0$ This is an autonomous separable ODE and a classic Bernoulli equation with power $n = 2$, characteristic of second-order bimolecular coagulation kinetics.

Common Errors: First-Order Equations

Error 1: Forgetting to normalize before identifying $P(x)$

In $a(x) y' + b(x) y = c(x)$, picking $P(x) = b(x)$ without first dividing through by the leading coefficient $a(x)$.

• Correction: The integrating factor requires canonical unit leading coefficient: $y' + \frac{b(x)}{a(x)} y = \frac{c(x)}{a(x)} \implies P(x) = \frac{b(x)}{a(x)}$

Error 2: Leaving the integration constant outside the reciprocal multiplier

Writing $y(x) = \frac{\int \mu Q\,dx}{\mu(x)} + C$ instead of the correct factored form: $y(x) = \frac{1}{\mu(x)} \left[\int \mu(x) Q(x)\,dx + C\right] = y_p(x) + C \mu(x)^{-1}$

• Consequence: The constant $C$ must decay along with the transient solution $e^{-\int P dx}$.

Error 3: Inverting cross-derivatives when testing exactness

In $M\,dx + N\,dy = 0$, evaluating $\partial_x M$ and $\partial_y N$ rather than the genuine mixed second partials $\partial_y M$ and $\partial_x N$.

• Mnemonic rule: $M$ accompanies $dx$, so test its sensitivity to the conjugate variable: $\partial_y M$.

Error 4: Discarding singular equilibrium roots during variable separation

In $\frac{dy}{dx} = y(1 - y)$, dividing through by $y(1 - y)$ without checking that $y(t) = 0$ and $y(t) = 1$ satisfy the ODE identically for all $t$.

Error 5: Neglecting the chain rule factor in Bernoulli transformations

Setting $u = y^{1-n}$ in $y' + P y = Q y^n$ but dropping the factor $(1-n)$ when replacing $y^{-n} y'$: $u' = (1-n) y^{-n} y' \implies y^{-n} y' = \frac{1}{1-n} u'$

Quick control rule

To verify your integrating factor $\mu(x)$: compute $\frac{d}{dx}(\mu(x) y) = \mu y' + \mu' y$. It must equal $\mu(x)(y' + P(x) y)$ identically.

Physical Applications: First-Order Equations

1. Timing circuits and passive RC low-pass filters

Capacitor charging through resistance underlies 555-timer relaxation oscillators, passive low-pass filters, and input anti-aliasing circuitry: $\frac{dV_C}{dt} + \frac{1}{R C} V_C = \frac{V_{\text{in}}(t)}{R C}$ The $-3\,\text{dB}$ cutoff frequency $f_c = \frac{1}{2\pi R C}$ filters out high-frequency electromagnetic interference in sensor telemetry and audio electronics.

2. Sensor thermal response and dynamic compensation

A thermocouple or RTD sensor plunged into a fluid does not respond instantaneously; it tracks according to: $\frac{dT_s}{dt} + \frac{h A}{m c_p} T_s = \frac{h A}{m c_p} T_f(t)$ where $\tau = \frac{m c_p}{h A}$ is the thermal response time constant. Correcting for this first-order lag is crucial in gas turbine pyrometry and fast-response environmental sensing.

3. Sedimentation dynamics and biological ultracentrifugation

The migration of macromolecules and organelles subjected to centrifugal fields obeys: $m \frac{dv}{dt} = m \omega^2 r \left(1 - \frac{\rho_{\text{fluid}}}{\rho_{\text{particle}}}\right) - 6 \pi \eta R_h v$ Rapidly reaching terminal velocity ($\frac{dv}{dt} \approx 0$), sedimenting velocity scales with the Svedberg unit ($S$), allowing separation of DNA, RNA, and protein complexes.

4. Pharmacokinetics and single-compartment clearance

Plasma concentration of a drug under continuous IV infusion at rate $R_0$ with first-order hepatic/renal elimination satisfies: $V_d \frac{dC}{dt} = R_0 - k_e V_d C \implies \frac{dC}{dt} + k_e C = \frac{R_0}{V_d}$ where $V_d$ is distribution volume and $k_e$ elimination clearance. Solving this ODE sets steady-state therapeutic dosing regimes in clinical pharmacology.

5. Radiative transfer and optical depth in astrophysics

Radiation transfer through stellar atmospheres or molecular clouds without emission follows: $\frac{dI_\nu}{ds} = -\kappa_\nu \rho I_\nu$ where $\kappa_\nu$ is spectral opacity. Integrating yields $I_\nu(s) = I_{\nu, 0} e^{-\tau_\nu}$, defining optical depth $\tau_\nu = \int \kappa_\nu \rho\,ds$, essential for determining exoplanet atmospheric compositions via transmission spectroscopy.

Interdisciplinary connection

First-order ODEs drive continuous stirred-tank reactors (CSTR) in chemical engineering, information cascade dynamics across social networks, and exponential skill decay/memory retention models in cognitive science.

Guided Practice: First-Order Equations

Exercise 1: Free fall with quadratic aerodynamic drag

A skydiver of mass $m = 80\,\text{kg}$ falls under gravity ($g = 9.8\,\text{m}/\text{s}^2$) against air resistance proportional to velocity squared: $F_d = -c v^2$ with $c = 0.2\,\text{kg}/\text{m}$.

1. Formulate the separable ODE for $v(t)$ starting from rest $v(0) = 0$.
2. Compute the terminal velocity $v_L$.
3. Solve for $v(t)$ analytically by separation of variables.

Solution

1. Dynamic ODE: $m \frac{dv}{dt} = m g - c v^2 \implies \frac{dv}{dt} = \frac{c}{m}(v_L^2 - v^2)$
2. Terminal velocity ($\frac{dv}{dt} = 0$): $v_L = \sqrt{\frac{m g}{c}} = \sqrt{\frac{80 \times 9.8}{0.2}} = \sqrt{3920} \approx 62.61\,\text{m}/\text{s} \approx 225.4\,\text{km}/\text{h}$
3. Separation of variables: $\frac{dv}{v_L^2 - v^2} = \frac{c}{m} dt$ Using partial fractions $\frac{1}{v_L^2 - v^2} = \frac{1}{2 v_L}\left(\frac{1}{v_L - v} + \frac{1}{v_L + v}\right)$: $\frac{1}{2 v_L} \ln\left|\frac{v_L + v}{v_L - v}\right| = \frac{c}{m} t + C_1$ With $v(0) = 0 \implies C_1 = 0$: $v(t) = v_L \tanh\left(\frac{g}{v_L} t\right) = 62.61 \tanh(0.1565 t)\,\text{m}/\text{s}$

Exercise 2: Charging a capacitor with linearly ramping voltage (Linear ODE)

An $RC$ circuit with $R = 1000\,\Omega$ and $C = 10\,\mu\text{F} = 10^{-5}\,\text{F}$ is powered by a ramp generator: $V(t) = 10 t\,\text{V}$. Given uncharged initial state $q(0) = 0\,\text{C}$:

1. Write the linear ODE for charge $q(t)$.
2. Calculate the integrating factor $\mu(t)$.
3. Obtain the exact closed-form expression for $q(t)$.

Solution

1. Kirchhoff loop ODE: $\frac{dq}{dt} + \frac{1}{R C} q = \frac{V(t)}{R} \implies \frac{dq}{dt} + 100 q = 0.01 t$
2. Integrating factor: $\mu(t) = \exp\left(\int 100\,dt\right) = e^{100 t}$
3. Integrating by parts: $q(t) e^{100 t} = 0.01 \int t e^{100 t}\,dt = 0.01 \left[\frac{t e^{100 t}}{100} - \frac{e^{100 t}}{10000}\right] + K$ $q(t) = 10^{-4} t - 10^{-6} + K e^{-100 t}$ Applying $q(0) = 0 \implies K = 10^{-6}\,\text{C}$: $q(t) = 10^{-4} t - 10^{-6}\left(1 - e^{-100 t}\right)\,\text{C}$

Exercise 3: Exact differential equation in thermodynamics

Consider the differential 1-form: $(2 x y + \cos x)\,dx + (x^2 + 3 y^2)\,dy = 0$

1. Prove exactness analytically via Euler-Schwarz symmetry.
2. Find the scalar potential function $\Phi(x, y)$ such that $d\Phi = 0$.
3. Determine the integral curve passing through $(0, 2)$.

Solution

1. Exactness test: $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2 x y + \cos x) = 2 x$ $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^2 + 3 y^2) = 2 x$ Both equal $2 x$, confirming exactness.
2. Integrating potential: $\Phi(x, y) = \int (2 x y + \cos x)\,dx + h(y) = x^2 y + \sin x + h(y)$ Matching $y$-derivative: $\frac{\partial \Phi}{\partial y} = x^2 + h'(y) = x^2 + 3 y^2 \implies h'(y) = 3 y^2 \implies h(y) = y^3$ Potential: $\Phi(x, y) = x^2 y + \sin x + y^3 = C$
3. Passing through $(0, 2)$: $\Phi(0, 2) = 0 + 0 + 8 = 8 \implies C = 8$ The solution is the implicit curve: $x^2 y + \sin x + y^3 = 8$