Differential equations

Every equation so far had a number for its unknown. A differential equation has a function for its unknown, and describes it by how it changes:

$$\frac{dy}{dx} = 2y$$

“Find the function whose rate of change is twice its value.” That is a description of behaviour, not of a value, which is why differential equations are how science states its laws — chapters 2.1 and 10.1 of PhysTB are both exactly this.

Checking a solution is easy

Solving is work; verifying is just differentiating.

Is $y = 3e^{2x}$ a solution of $\tfrac{dy}{dx} = 2y$?

Both sides come to $6e^{2x}$, so yes. Note that $5e^{2x}$ works too, and so does $Ce^{2x}$ for any constant — a differential equation has a family of solutions, and pinning down which one needs an extra fact.

That extra fact is an initial condition like $y(0) = 3$, and a problem giving you one is asking for a single function rather than a family.

Separation of variables

The one technique AB needs. When the equation can be written with all the $y$s on one side and all the $x$s on the other, integrate both sides.

For $\tfrac{dy}{dx} = 2y$:

$$\frac{dy}{y} = 2\,dx \quad\Longrightarrow\quad \int\frac{dy}{y} = \int 2\,dx \quad\Longrightarrow\quad \ln|y| = 2x + C$$

Exponentiating gives $y = Ce^{2x}$, and $y(0) = 3$ makes $C = 3$.

Where the constant goes. One constant is enough — combining the two from the separate integrals gives a single $C$, and exponentiating turns an additive constant into a multiplicative one. That is why the answer has $C$ as a coefficient rather than an addend.

Exponential growth

$\tfrac{dy}{dt} = ky$ says the rate of growth is proportional to the amount present, and its solution is

$$y = y_0 e^{kt}$$

Positive $k$ is growth, negative is decay — the second claim is the decay case heading to zero. This one equation covers compound interest, population growth without constraint, and radioactive decay.

Doubling and half-life. Time to double satisfies $e^{kt} = 2$, so $t = \tfrac{\ln 2}{k}$ — independent of where you started, which is the characteristic property of exponential behaviour and a standard exam question.

Logistic growth

Unconstrained exponential growth is unphysical: populations run out of food. The logistic equation adds a ceiling $L$:

$$\frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right)$$

The bracket is near $1$ when $y$ is small — so growth starts exponential — and near $0$ as $y$ approaches $L$, which stalls it. The population levels off at $L$, the carrying capacity.

Two facts the exam asks for directly, and both come from the equation without solving it:

• $\lim_{t\to\infty} y = L$. Growth stops when the bracket vanishes.
• Growth is fastest at $y = L/2$. The rate $ky(1 - y/L)$ is a downward parabola in $y$, and its vertex is halfway.

Setting that derivative to zero gives $y = \tfrac{L}{2}$ exactly — the maximum growth rate, verified. That point is the inflection point of the solution curve, where it stops curving up and starts curving down.

Slope fields

A slope field draws a short segment at each point with the slope the equation prescribes there. Solution curves follow the segments, and one passes through each point — which is a picture of the family of solutions all at once.

Reading one:

• Follow the segments from the initial condition to sketch that solution.
• Horizontal segments mark where $\tfrac{dy}{dx} = 0$ — equilibria.
• Segments depending only on $y$ mean the field looks the same at every $x$, which is the signature of an autonomous equation like $\tfrac{dy}{dt} = ky$.

For the logistic equation the field is horizontal along $y = 0$ and $y = L$ — the two equilibria — steepest in between, and every curve starting between them climbs toward $L$.