Differential equations, and separable equations

What a differential equation is

A differential equation is an equation for an unknown function that involves its derivative. In $y' = 0.5y$ the unknown is a function $y(t)$. The equation says that at every time $t$, the slope of $y$ is half its value.

With one input variable, as here, it is an ordinary differential equation, or ODE. Module 6.3 meets equations for functions of two variables.

Units follow the derivative. If $y$ is in grams and $t$ in hours, then $y' = \frac{dy}{dt}$ is in grams per hour, and 0.5 is a rate per hour. So $y$ grows by 50% of its size per hour: its relative rate (lesson 2.3.4).

The order of an equation is its highest derivative. $y' = 0.5y$ and $y' = t - y$ are first order; $y'' = -y$ is second order, the subject of module 6.2. This module treats first-order equations written as $y' = f(t, y)$: the slope is a given function of the time and the current value.

The simplest have no $y$ on the right. $y' = 2t$ is solved by every antiderivative $t^2 + C$ (lesson 2.5.2). A differential equation is integration in which the slope may also depend on the answer.

What counts as a solution

A solution is a function that makes the equation true at every $t$ of an interval. To check a candidate, substitute it: work out both sides and compare them.

Three candidates for $y' = 0.5y$:

• $y = 3e^{0.5t}$ has $y' = 1.5e^{0.5t}$, and $0.5y = 1.5e^{0.5t}$. They agree for every $t$: a solution.
• $y = e^{0.5t} + 1$ has $y' = 0.5e^{0.5t}$, but $0.5y = 0.5e^{0.5t} + 0.5$. The two sides always differ by 0.5: not a solution.
• $y = t^2$ has $y' = 2t$ and $0.5y = 0.5t^2$. These agree only at $t = 0$ and $t = 4$. Two points are not enough: not a solution.

Initial conditions

Every function $Ce^{0.5t}$ solves $y' = 0.5y$, whatever the constant $C$. The solutions form a family, as the antiderivatives $t^2 + C$ did in lesson 2.5.2. One known value picks one member.

A known value $y(t_0) = y_0$ is an initial condition, and the equation with it is an initial value problem. With $y(0) = 3$, $C = 3$. With $y(2) = 5$ instead, $Ce^{1} = 5$, so $C = \frac{5}{e} \approx 1.839397$.

When $f$ and its partial derivative $\frac{\partial f}{\partial y}$ (lesson 4.1.2) are continuous, exactly one solution passes through each starting point, at least for a while. So two solutions of the same equation never cross.

The slope field

The equation $y' = f(t, y)$ gives the slope of the solution through any point $(t, y)$, before any solution is known. Draw a short line with that slope at each point of a grid: this is the slope field. A solution is a curve that runs along the lines everywhere.

For $y' = 0.8(2 - y)$, the slope at $(0, 4)$ is $0.8 \times (-2) = -1.6$, at $(0, 2)$ it is 0, and at $(0, 0)$ it is 1.6. Here the slope depends only on $y$, so every column of the field looks the same.

Equilibrium solutions

If $f(t, c) = 0$ for every $t$, the constant $y = c$ is a solution, with slope 0 throughout. It is an equilibrium. For $y' = 0.8(2 - y)$ it is $y = 2$, and for $y' = 0.5y$ it is $y = 0$. The logistic equation $y' = y\left(1 - \frac{y}{4}\right)$ has two: $y = 0$ and $y = 4$.

An equilibrium is stable when nearby solutions move towards it and unstable when they move away. Above 2, $0.8(2 - y)$ is negative, so $y$ falls; below 2 it rises. So 2 is stable, while for $y' = 0.5y$ a positive $y$ grows and a negative one falls: 0 is unstable.

Separable equations

An equation is separable when its slope is a function of $t$ times a function of $y$: $\frac{dy}{dt} = g(t)\,h(y)$. Then the $y$ terms can go to one side and the $t$ terms to the other, and each side is integrated in its own variable.

Moving $dt$ across is shorthand, as $du = 2x\,dx$ was in lesson 2.6.1. If $H$ is an antiderivative of $\frac{1}{h}$, the chain rule gives $\frac{d}{dt}H(y(t)) = \frac{y'}{h(y)} = g(t)$, and integrating that in $t$ is the rule.

Exponential growth and decay

Take $y' = ky$ with a constant $k$, so $g(t) = k$ and $h(y) = y$.

1. 1
   $$\frac{dy}{y} = k\,dt$$

   Why?
2. 2
   $$\ln|y| = kt + C$$

   Why?
3. 3
   $$y = A\,e^{kt}$$

   Why?

At $t = 0$, $y = A$, so $A$ is the starting value $y_0$.

For $y' = 0.5y$ with $y(0) = 1$, each unit of time multiplies $y$ by $e^{0.5} \approx 1.648721$:

It doubles every $\frac{\ln 2}{0.5} \approx 1.3863$ units of time. Decay at $k = -0.2$ halves the size every $\frac{\ln 2}{0.2} \approx 3.4657$ units.

Settling on an equilibrium

$y' = 0.8(2 - y)$ separates with $g(t) = 0.8$ and $h(y) = 2 - y$. Integrate $\frac{1}{2 - y}$ with the linear substitution $u = 2 - y$, $du = -dy$ (lesson 2.6.1). Then raise $e$ to both sides, as above, and let $y(0) = y_0$ fix the constant:

The gap to the equilibrium, $y - 2$, decays at the rate 0.8. Each unit of time multiplies it by $e^{-0.8} \approx 0.4493$, and it halves every $\frac{\ln 2}{0.8} \approx 0.8664$.

These are the curves in the first panel of the figure, with $k = 0.8$ and $M = 2$:

When $y$ is not exponential

Not every separable equation gives an exponential. Take $y' = \frac{t}{y}$ with $y(0) = 2$. Separating gives $y\,dy = t\,dt$, so $\frac{y^2}{2} = \frac{t^2}{2} + C$ and $y^2 = t^2 + 4$ once $y(0) = 2$ fixes $C$.

Two functions satisfy that: $\sqrt{t^2 + 4}$ and $-\sqrt{t^2 + 4}$. The start $y(0) = 2$ is positive, so the solution is $y = \sqrt{t^2 + 4}$; from $y(0) = -2$ it would be the negative root. Substituting checks it: $y' = \frac{t}{\sqrt{t^2 + 4}} = \frac{t}{y}$. At $t = 1, 2, 3, 4$ it is 2.236068, 2.828427, 3.605551 and 4.472136.

Solutions that blow up

$y' = y^2$ with $y(0) = 1$ looks harmless. Separating gives $-\frac{1}{y} = t + C$, so $C = -1$ and $y = \frac{1}{1 - t}$. At $t = 0.9$, 0.99 and 0.999 it is 10, 100 and 1000: it blows up at $t = 1$ (lesson 2.1.2).

So a solution may exist only on an interval, here $t < 1$, even when the equation itself is defined everywhere. Dividing by $y^2$ also lost the equilibrium $y = 0$.

Equations that do not separate

$y' = t + y$ is a sum, not a product $g(t)\,h(y)$, and no rearranging makes it one. It is a linear equation, which lesson 6.1.2 solves with an integrating factor.

Euler's method

Most differential equations have no solution in familiar functions, but every one has a slope field to follow. Euler's method follows it in straight steps of length $h$ in $t$. From the start, move along the line with the slope $f(t_0, y_0)$ for one step, then read the slope again.

For $y' = 0.5y$ with $y(0) = 1$ and $h = 0.5$, each step multiplies $y$ by $1 + 0.5h = 1.25$. The exact solution multiplies it by $e^{0.25} \approx 1.284025$ over the same step, so Euler falls behind:

The errors are negative because the curve bends up, so each tangent line lies below it (lesson 2.4.2). The shortfalls build up from step to step.

The error is proportional to $h$

At $t = 2$ the error is $-0.064984$ with $h = 0.1$, $-0.033218$ with $h = 0.05$ and $-0.000679$ with $h = 0.001$. Halving the step about halves the error: Euler's method is first order, like the forward difference of lesson 2.8.1. Even 2,000 steps leave an error of 0.000679; lesson 6.1.2's solve_ivp does far better.

A step that is too long can overshoot. For $y' = 0.8(2 - y)$, each Euler step multiplies the gap $y - 2$ by $1 - 0.8h$. With $h = 1.5$ that factor is $-0.2$: from 4 the first step lands on 1.6, below the equilibrium, then 2.08 and 1.984. With $h = 3$ it is $-1.4$, and the steps run 4, $-0.8$, 5.92, $-3.488$: the numbers blow up while the true solution settles.

Below, the dark curve is the exact solution of $y' = 0.8(2 - y)$ and the coloured path is Euler's method. Move the start above and below 2, then compare $h = 1$ with $h = 0.1$: from $y(0) = 4$, the error at $t = 5$ shrinks from $-0.036$ to $-0.0057$. Set $h = 1.5$ to watch each step jump past the equilibrium.

The buttons at the top switch to $y' = 0.5y$, to the logistic equation and to an equation with a rate that rises with $t$. Case study 6.1.4 uses that last one to price a bond.

Common mistakes

1. Check a solution at every $t$, not at one point. $t^2$ satisfies $y' = 0.5y$ at $t = 0$ and $t = 4$ only.
2. The constant multiplies. $\ln|y| = kt + C$ gives $y = Ae^{kt}$, not $e^{kt} + C$. Fix it with the initial condition.
3. Dividing by $h(y)$ loses the equilibria. $y = 0$ solves $y' = y^2$, though separating never finds it.
4. Let the initial condition choose the root. From $y(0) = 2$, $y' = \frac{t}{y}$ gives $\sqrt{t^2 + 4}$, not $-\sqrt{t^2 + 4}$.
5. A solution may stop. $\frac{1}{1 - t}$ solves $y' = y^2$ only for $t < 1$.
6. Not every equation separates. $y' = t + y$ is a sum; it needs lesson 6.1.2.
7. Euler's error shrinks only in proportion to $h$, and too long a step can overshoot an equilibrium or blow up.