Differential Equations

A differential equation (DE) is an equation involving an unknown function and its derivatives, such as $\dfrac{dy}{dx}=2y$ or $\dfrac{dP}{dt}=kP$. A solution is a function $y=f(x)$ that makes the equation true when substituted in. The general solution contains an arbitrary constant $C$ (a whole family of curves); a particular solution is pinned down by an initial condition like $y(0)=3$.

A first-order DE $\dfrac{dy}{dx}=F(x,y)$ gives the slope of the solution curve at every point. A slope field draws a short segment with that slope at a grid of points. To sketch a solution through a given point, start there and “flow” along the segments, always keeping the curve tangent to the nearby marks.

A DE is separable if it can be written $\dfrac{dy}{dx}=g(x)\,h(y)$. Separate the variables so each side involves only one letter, then integrate:

Add a single constant $+C$, then apply the initial condition to solve for $C$ (an initial-value problem, IVP).

The single most important DE in the course is

where $y_0=y(0)$ is the initial amount. $k>0$ gives growth; $k<0$ gives decay.

When growth is limited by a carrying capacity $M$, use

When $P$ is small, $1-\frac{P}{M}\approx1$, so growth is nearly exponential; as $P\to M$ the factor $\to0$ and growth stalls.

When a DE cannot be solved in closed form, approximate the solution numerically. Starting from $(x_0,y_0)$ with step size $h$, repeatedly follow the tangent line:

Each step uses the slope $F$ at the current point to predict the next $y$.

Let $y(t)$ be the amount of a substance in a tank. Then

where each rate is (concentration)$\times$(flow). If the tank is well-mixed, the outflow concentration is $\dfrac{y}{V}$ (current amount over current volume).

Differentiate $\dfrac{dP}{dt}=kP\!\left(1-\frac{P}{M}\right)$ with respect to $t$ to study $\dfrac{d^2P}{dt^2}$. It is positive for $0<P<\frac M2$ (curve concave up, growth accelerating) and negative for $\frac M2<P<M$ (concave down, growth decelerating). The sign change at $P=\frac M2$ is the inflection point --- the moment of fastest growth --- which is why the graph rises, bends, and levels off toward $M$.

Euler's method is first order: the accumulated (global) error is roughly proportional to the step size $h$. Halving $h$ roughly halves the error but doubles the number of steps. The local error each step comes from ignoring curvature ($y''$), so more curvature means faster error growth. This is why the concavity of the true solution tells you the direction of the error.

An equilibrium solution satisfies $F(x,y)=0$ (a constant solution). Classify it from the sign of $\dfrac{dy}{dx}$ nearby:

• [leftmargin=*]
• Stable (a sink): solutions on both sides move toward it. For logistic growth, $P=M$ is stable.
• Unstable (a source): nearby solutions move away. For logistic growth, $P=0$ is unstable.

A phase line records these arrows and predicts long-term behavior without solving the DE.

Higher-order DEs involve $y''$. For example, $y''=-\omega^2 y$ models simple harmonic motion, with general solution

Two arbitrary constants $A,B$ appear because the equation is second order; two conditions (e.g. $y(0)$ and $y'(0)$) pin down a particular solution. You will meet these systematically in a later course.