Separable equations

The previous lesson could draw the family of solutions to a first order equation without solving it, and this one solves a genuine class of them outright.

The class is narrow but it contains most of the standard models: radioactive decay, drug clearance, cooling, dissolution, and population growth against a limit. What they share is that the rate of change factors, with everything depending on time in one bracket and everything depending on the unknown in another. When that happens, calculus alone finishes the job.

The factoring that makes integration possible

An equation is separable if it can be written

\frac{d y}{d t} = g (t) h (y)

The method is to divide by $h (y)$ , multiply by $d t$ , and integrate both sides. That description is a mnemonic rather than an argument, since $d t$ is not a quantity to be multiplied by, so here is the argument. Divide by $h (y)$ to get

\frac{1}{h (y)} \frac{d y}{d t} = g (t)

and integrate both sides with respect to $t$ . On the right that gives $\int g (t) d t$ . On the left, the integrand is exactly of the form $F' (y) y'$ with $F$ an antiderivative of $1 / h$ , so the chain rule identifies it as the derivative of $F (y (t))$ , and by substitution $\int \frac{1}{h (y)} \frac{d y}{d t} d t = \int \frac{d y}{h (y)}$ . So

\int \frac{d y}{h (y)} = \int g (t) d t + C

which is the mnemonic, now justified. Only one constant is needed, since the two would merge. What comes out is a relation between $y$ and $t$ , not necessarily a formula for $y$ , and that distinction matters below.

Example. Solve $\frac{d y}{d t} = t y$ with $y (0) = 3$ .

Separating, $\frac{d y}{y} = t d t$ , so $\ln | y | = {\frac{12}{t}}^{2} + C$ and $| y | = e^{C} e^{t^{2} / 2}$ . Absorbing the sign and the exponential of the constant into one constant $A$ , the general solution is $y = A e^{t^{2} / 2}$ . The initial condition gives $A = 3$ , so $y = 3 e^{t^{2} / 2}$ . Checking: $y' = 3 t e^{t^{2} / 2} = t y$ , as required.

Now you. Solve $\frac{d y}{d t} = y \cos t$ with $y (0) = 2$ .

Answer

$\frac{d y}{y} = \cos t d t$ gives $\ln | y | = \sin t + C$ , so $y = A e^{\sin t}$ and $A = 2$ . The solution is $y = 2 e^{\sin t}$ , which oscillates between $2 e^{- 1} = 0.736$ and $2 e = 5.44$ .

The solutions that division destroys

Dividing by $h (y)$ is only legal where $h (y)$ is not zero, and every root of $h$ is a constant solution that the method throws away. If $h (y_{0}) = 0$ then the constant function $y \equiv y_{0}$ has zero derivative and satisfies the equation exactly, yet it can never appear in the family produced by separating, because the very first step divided by zero to get there.

For $y' = y (1 - y)$ the roots are $y = 0$ and $y = 1$ , and both are genuine solutions that must be added back by hand. Sometimes the lost solution reappears as a limiting case of the constant: for $y' = k y$ the family $y = A e^{k t}$ includes $y \equiv 0$ at $A = 0$ , even though the derivation assumed $y \neq 0$ . Sometimes it does not, and then the general solution is genuinely incomplete without it. Check the roots of $h$ every time; it costs one line and it is the standard way to lose half an answer.

Decay, growth and half-lives

The equation $\frac{d y}{d t} = k y$ says the rate of change is proportional to the amount present, and separating gives $y = y_{0} e^{k t}$ . Negative $k$ is decay, positive is unconstrained growth. What makes the model useful is that $k$ is rarely quoted directly: what is measured is a half-life $t_{1 / 2}$ , the time for the amount to halve, and setting $y = y_{0} / 2$ gives $e^{k t_{1 / 2}} = \frac{12}{}$ , so $k = - \ln 2 / t_{1 / 2}$ .

Caffeine leaves the bloodstream by first order kinetics with a half-life near $5$ hours in a typical adult. So $k = - \ln 2 / 5 = - 0.1386$ per hour, and a $200$ mg dose falls to $95$ mg after $t = \ln (200 / 95) / 0.1386 = 5.4$ hours. The same arithmetic dates a bone from its carbon-14, sizes a shielding delay from a reactor isotope, and sets the dosing interval of a drug. The characteristic quantity is $1 / | k |$ , the time constant, after which the amount has fallen by a factor of $e$ ; a half-life is $\ln 2 = 0.693$ time constants.

The honest limit is that pure exponential growth is a statement about a system with unlimited resources, and no real population has those. It is the first term of a better model, not the model.

Newton's law of cooling, fitted to a real mug

A body at temperature $T$ in surroundings at $T_{s}$ loses heat at a rate roughly proportional to the excess temperature, which gives

\frac{d T}{d t} = - k (T - T_{s})

Separating with $u = T - T_{s}$ , which has the same derivative as $T$ since $T_{s}$ is constant, gives $u = u_{0} e^{- k t}$ , so

T (t) = T_{s} + (T_{0} - T_{s}) e^{- k t}

Every cooling curve is the same shape: the excess over ambient decays exponentially, and the temperature approaches the room's, never crossing it.

Example. Coffee at $90$ °C stands in a room at $20$ °C. After $5$ minutes it reads $70$ °C. When will it reach $45$ °C?

The excess starts at $70$ degrees and is $50$ degrees after five minutes, so $50 = 70 e^{- 5 k}$ and $k = \frac{15}{\ln} (70 / 50) = 0.0673$ per minute. Drinking temperature at $45$ °C is an excess of $25$ degrees, so $25 = 70 e^{- k t}$ and $t = \ln (70 / 25) / 0.0673 = 15.3$ minutes. As a check, the prediction at ten minutes is $20 + 70 e^{- 0.673} = 55.7$ °C, which is the same fall of $70$ to $50$ per cent applied twice.

Now you. Soup at $95$ °C sits in a room at $22$ °C and is at $65$ °C after $10$ minutes. When does it reach $40$ °C?

Answer

The excess falls from $73$ to $43$ in ten minutes, so $k = \frac{1}{10} \ln (73 / 43) = 0.0529$ per minute. For an excess of $18$ degrees, $t = \ln (73 / 18) / 0.0529 = 26.5$ minutes.

The law is an approximation, valid when the excess temperature is small enough that radiation, which goes as the fourth power of absolute temperature, is not dominant, and when the body is well stirred or thin enough to have one temperature at all. For a mug of coffee at kitchen temperatures both hold well enough that the fit above is good to a degree or so over half an hour.

Implicit solutions and the interval they live on

Separation produces a relation, and there is no guarantee that the relation can be solved for $y$ . Take $\frac{d y}{d t} = - \frac{t}{y}$ with $y (0) = 3$ . Separating gives $y d y = - t d t$ , so ${\frac{12}{y}}^{2} = - {\frac{12}{t}}^{2} + C$ and

t^{2} + y^{2} = 9

a circle. That is a perfectly good implicit solution, and it says more clearly than any formula what the solutions look like. Made explicit, $y = \sqrt{9 - t^{2}}$ , taking the positive root because $y (0) = 3$ , and now the trouble is visible: the solution exists only for $- 3 < t < 3$ . At the ends the circle has a vertical tangent, the derivative is infinite, and the solution simply stops. Nothing in the original equation announced the number $3$ ; it came from the initial condition. The interval of existence of a solution can depend on where the solution starts, which never happens for the linear equations of the next lesson.

Some separable equations lead to integrals that need the techniques of the previous course. $\frac{d y}{d t} = \frac{y^{2} - 1}{2}$ separates to $\int \frac{2 d y}{(y - 1) (y + 1)} = \int d t$ , a partial fractions problem, and others lead to integrals with no elementary form at all, at which point separation has converted a differential equation into a definite integral to be done numerically. That is still progress.

The logistic equation

Exponential growth fails because it ignores the resource limit. Model the limit in the crudest way that could work: let the per-capita growth rate fall linearly to zero as the population $P$ approaches a carrying capacity $K$ . The per-capita rate is $\frac{1}{P} \frac{d P}{d t}$ , so the assumption is $\frac{1}{P} \frac{d P}{d t} = r (1 - \frac{P}{K})$ , giving the logistic equation

\frac{d P}{d t} = r P (1 - \frac{P}{K})

published by Pierre-François Verhulst in 1838. It is separable and nonlinear. Separating and using partial fractions on $\frac{1}{P (1 - P / K)} = \frac{1}{P} + \frac{1 / K}{1 - P / K}$ gives $\ln P - \ln (1 - P / K) = r t + C$ , so $\frac{P}{1 - P / K} = A e^{r t}$ , and solving for $P$ ,

P (t) = \frac{K}{1 + (\frac{K}{P_{0}} - 1) e^{- r t}}

The behaviour is the S-curve: growth is nearly exponential at rate $r$ while $P ≪ K$ , slows as the bracket shrinks, and levels off at $K$ . The steepest point is where $\frac{d P}{d t}$ is largest, and since that is a downward parabola in $P$ with roots at $0$ and $K$ , the maximum is at $P = K / 2$ , exactly half the capacity, at which the growth rate is $r K / 4$ .

Example. A lake is stocked with $100$ fish. The population grows logistically with $r = 0.03$ per year and $K = 1000$ . When does it reach $500$ ?

Half capacity is reached when the denominator is $2$ , that is when $(\frac{1000}{100} - 1) e^{- r t} = 1$ , so $9 e^{- 0.03 t} = 1$ and $t = \ln 9 / 0.03 = 73.2$ years. That is also the moment of fastest growth, at $r K / 4 = 7.5$ fish per year.

Now you. A population starts at $50$ with $r = 0.05$ per year and $K = 800$ . When does it reach $400$ ?

Answer

$(\frac{800}{50} - 1) e^{- 0.05 t} = 1$ gives $15 e^{- 0.05 t} = 1$ and $t = \ln 15 / 0.05 = 54.2$ years.

Raymond Pearl and Lowell Reed fitted a logistic to the United States census in 1920 and obtained $K = 197.3$ million, $r = 0.0313$ per year, and half capacity in 1914. Against the census the fit is remarkable: it predicts $77.3$ million for 1900 against a measured $76.2$ , and $107.9$ million for 1920 against $106.0$ . It is also wrong in the way models of this sort are always wrong. The United States passed $197$ million in 1967 and kept going, because immigration and changing fertility are not in the model, and by 2020 the fit predicts $190$ million against an actual $331$ . A logistic fit measures the growth pattern of the period it was fitted to, and its extrapolated $K$ is a summary of that period rather than a property of the country.

Where separation runs out

Separation needs the right side to factor, and most equations do not. Consider a stirred tank whose inflow concentration varies during the day, or a capacitor charged from a mains supply. Both give equations of the shape

\frac{d y}{d t} + p (t) y = g (t)

with $g$ not constant, and there is no way to write $g (t) - p (t) y$ as a product of a function of $t$ and a function of $y$ . Even the simplest instance, $y' = t + y$ , resists. Yet these equations are linear, which the logistic equation is not, and linear equations are supposed to be the easy ones.

They are, once the right trick is found. The next lesson finds it by asking what could be multiplied through the equation to make the left side the derivative of a single product, and the answer solves every first order linear equation there is.