Differential equations and dynamical systems

First-order differential equations

Equations relating a function to its first derivative, solved by separation of variables or integrating factors.

IntuitionHow fast is it changing right now?

Many phenomena are easiest to describe not by a formula for the quantity itself, but by a rule for how fast that quantity is changing: a population grows in proportion to its current size, a hot cup of coffee cools at a rate proportional to how much hotter it is than the room, a charging capacitor's voltage rises more slowly the closer it gets to full. A first-order differential equation is exactly such a rule — it relates an unknown function $y(x)$ to its own derivative $y'(x)$ — and solving it means recovering the full function from the rule for its rate of change.

Slope field of a first-order ODE

y' = y - c x^2

with three solution curves starting from different initial values

y(t_0) = y_0

UndergraduateSeparable and linear equations

Definition: Separable equation

A first-order equation is separable when it can be written as $\frac{dy}{dx}=f(x)g(y)$ : the right-hand side factors into a part depending only on $x$ and a part depending only on $y$ . Moving all $y$ -terms to one side and all $x$ -terms to the other turns the equation into two ordinary integrals, $\int \frac{dy}{g(y)}=\int f(x)\,dx$ .

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

For instance, the autonomous equation $\frac{dy}{dx}=g(y)$ (where the rate of change depends only on the current value $y$ , not on $x$ ) is the special case $f(x)=1$ ; its equilibrium solutions occur exactly where $g(y)=0$ , and these constant solutions are easy to lose if one divides by $g(y)$ carelessly.

\frac{dy}{dx}+P(x)y=Q(x)

Types of first-order equations
Type	Standard form	Solution method
Separable	$\frac{dy}{dx}=f(x)g(y)$	integrate both sides: $\int \frac{dy}{g(y)}=\int f(x)\,dx$
Linear first-order	$\frac{dy}{dx}+P(x)y=Q(x)$	multiply by $\mu(x)=e^{\int P(x)\,dx}$ , then integrate
Autonomous	$\frac{dy}{dx}=g(y)$	special separable case; equilibria at $g(y)=0$

UndergraduateSolving and guaranteeing a solution exists

Theorem: The integrating factor formula

If $P(x)$ and $Q(x)$ are continuous on an interval and $\mu(x)=e^{\int P(x)\,dx}$ , then every solution of the linear equation $\frac{dy}{dx}+P(x)y=Q(x)$ is given by $y(x) = \frac{1}{\mu(x)}\left(\int \mu(x) Q(x)\,dx + C\right)$ .

Why is it true?

The trick is analogous to completing the square: we multiply the whole equation by a cleverly chosen factor $\mu(x)$ so that the left-hand side collapses into the derivative of a single product $\mu(x)y(x)$ , turning a differential equation into something we can integrate directly.

Proof

Define $\mu(x)=e^{\int P(x)\,dx}$ ; differentiating with the chain rule gives $\mu'(x)=P(x)\mu(x)$ , since the derivative of the exponent $\int P(x)\,dx$ is $P(x)$ .

Multiply both sides of $\frac{dy}{dx}+P(x)y=Q(x)$ by $\mu(x)$ : this gives $\mu(x)y' + \mu(x)P(x)y = \mu(x)Q(x)$ .

By the product rule, $\frac{d}{dx}[\mu(x)y(x)] = \mu(x)y'(x) + \mu'(x)y(x)$ , and since $\mu'(x)=P(x)\mu(x)$ , this equals exactly $\mu(x)y' + \mu(x)P(x)y$ — the left-hand side above. So the equation becomes $\frac{d}{dx}[\mu(x)y(x)] = \mu(x)Q(x)$ .

Integrating both sides with respect to $x$ gives $\mu(x)y(x) = \int \mu(x)Q(x)\,dx + C$ for an arbitrary constant $C$ . Dividing through by $\mu(x)$ (which is never zero, being an exponential) yields $y(x) = \frac{1}{\mu(x)}\left(\int \mu(x) Q(x)\,dx + C\right)$ .

Theorem: The Picard–Lindelöf existence-uniqueness theorem

Suppose $f(x,y)$ is continuous on a rectangle around $(x_0,y_0)$ and satisfies a Lipschitz condition in $y$ : $|f(x,y_1)-f(x,y_2)|\le L|y_1-y_2|$ for some constant $L$ . Then the initial value problem $y'=f(x,y),\ y(x_0)=y_0$ has a unique solution on some interval containing $x_0$ .

Why is it true?

The Lipschitz condition bounds how steeply $f$ can vary in the $y$ -direction, which prevents nearby solution trajectories from splitting apart or crossing: it is exactly the condition needed to guarantee a well-defined flow, with precisely one trajectory passing through each point.

Proof

Rewriting the initial value problem as an integral equation, a continuous function $y(x)$ solves $y'=f(x,y),\ y(x_0)=y_0$ if and only if it solves $y(x) = y_0 + \int_{x_0}^x f(t,y(t))\,dt$ ; this equivalence follows from the fundamental theorem of calculus.

Define the Picard iteration operator $T$ on continuous functions by $T[y](x) = y_0 + \int_{x_0}^x f(t,y(t))\,dt$ , restricted to a small interval $I=[x_0-h,x_0+h]$ chosen so that $f$ stays in the rectangle where the hypotheses hold.

For two continuous functions $y_1,y_2$ on $I$ , the Lipschitz condition gives $|T[y_1](x)-T[y_2](x)| \le \int_{x_0}^x |f(t,y_1(t))-f(t,y_2(t))|\,dt \le L\,h\,\sup_{t\in I}|y_1(t)-y_2(t)|$ . Choosing $h$ small enough that $Lh<1$ makes $T$ a contraction mapping in the supremum norm.

By the Banach fixed-point theorem, a contraction on a complete metric space (the continuous functions on $I$ with the supremum norm) has exactly one fixed point $y$ , satisfying $y=T[y]$ . This fixed point is precisely the unique solution of the original initial value problem on $I$ .

UndergraduateReal-World Applications and Worked Examples

First-order equations model population growth capped by limited resources (logistic growth in ecology and epidemiology), charging and discharging RC circuits in electronics, Newton's law of cooling in thermodynamics, radioactive decay in physics and carbon dating, and continuously compounding interest with a time-varying rate in finance. The two worked examples below solve the most common patterns end to end.

Example: Logistic population growth

A population $P(t)$ grows according to the logistic model $\frac{dP}{dt}=rP\left(1-\frac{P}{K}\right)$ with initial population $P(0)=P_0$ . Find $P(t)$ .

Solution

This equation is separable: it rearranges to $\frac{dP}{P(1-P/K)}=r\,dt$ .

The partial fraction decomposition $\frac{1}{P(1-P/K)}=\frac1P+\frac{1/K}{1-P/K}$ lets us integrate the left side term by term, giving $\ln|P|-\ln|1-P/K|=rt+C_1$ , i.e. $\ln\left|\frac{P}{K-P}\right|=rt+C_2$ after combining constants and multiplying by $K$ .

Exponentiating gives $\frac{P}{K-P}=Ae^{rt}$ for a constant $A$ ; evaluating at $t=0$ with $P(0)=P_0$ gives $A=\frac{P_0}{K-P_0}$ .

Solving for $P$ and simplifying produces the closed form $P(t) = \frac{K P_0 e^{rt}}{K + P_0(e^{rt}-1)}$ , which starts near $P_0$ and approaches the carrying capacity $K$ as $t\to\infty$ , exactly the S-shaped curve seen in real population data.

Example: Charging an RC circuit

A capacitor charging through a resistor from a constant voltage source $V$ satisfies $\frac{dQ}{dt} + \frac{Q}{RC} = \frac{V}{R}$ with $Q(0)=0$ , where $Q(t)$ is the charge at time $t$ . Find $Q(t)$ .

Solution

This is linear with $P(t)=\frac{1}{RC}$ and source term $\frac{V}{R}$ , so the integrating factor is $\mu(t)=e^{t/(RC)}$ .

Multiplying through gives $\frac{d}{dt}\left[e^{t/(RC)}Q\right] = \frac{V}{R}e^{t/(RC)}$ ; integrating both sides yields $e^{t/(RC)}Q = VC\,e^{t/(RC)} + C_1$ for a constant $C_1$ .

Dividing by $e^{t/(RC)}$ gives $Q(t) = VC + C_1 e^{-t/(RC)}$ . Applying $Q(0)=0$ gives $C_1=-VC$ .

So $Q(t)=CV\left(1-e^{-t/(RC)}\right)$ : the charge rises from $0$ toward its final value $CV$ , with the approach governed by the time constant $RC$ — slower for a larger resistor or capacitor.

What is the integrating factor for the linear first-order equation $\frac{dy}{dx}+P(x)y=Q(x)$ ?

In the logistic growth model $\frac{dP}{dt}=rP\left(1-\frac{P}{K}\right)$ , what does the constant $K$ represent?

Solve the separable equation $\frac{dy}{dx}=xy$ with $y(0)=1$ .

Which condition guarantees the Picard–Lindelöf theorem gives a unique solution to $y'=f(x,y),\ y(x_0)=y_0$ ?

References

William E. Boyce, Richard C. DiPrima, Douglas B. Meade (2017). Elementary Differential Equations and Boundary Value Problems
Morris Tenenbaum, Harry Pollard (1985). Ordinary Differential Equations