Practice / Calculus

Separable differential equations: growth, decay, cooling and the logistic curve

Ten separable differential equations: exponential growth with a negative start, y′ = x/y, a solution that blows up in finite time, the equation whose solution is softplus, half-life, Newton's law of cooling, the sigmoid as the solution of y′ = y(1 − y), gradient flow on a quadratic loss, a constant solution lost by dividing, and a draining tank, with worked solutions and the mistakes of adding the constant too late, splitting a square root and forgetting where a solution stops.

Before you start

A differential equation describes how a quantity changes, and solving it recovers the quantity itself. When the rate factors into a function of $x$ times a function of $y$ the variables can be separated and each side integrated on its own. The sigmoid, softplus and the exponential decay of gradient flow on a quadratic loss all come out of equations like this. The work goes wrong in a few places: the constant of integration added after exponentiating instead of before, $\sqrt{x^2 + 9}$ split into $x + 3$ a constant solution thrown away by dividing by zero, the shift in Newton's law of cooling forgotten, and a formula used past the point where the solution stops. These ten problems cover each of them.

Separable. $\dfrac{dy}{dx} = g(x)\,h(y)$ Divide by $h(y)$ multiply by $dx$ and integrate both sides: $\displaystyle\int \frac{dy}{h(y)} = \int g(x)\,dx + C$ One constant is enough, and it goes in before you solve for $y$
Constant solutions. If $h(y_0) = 0$ then $y \equiv y_0$ is a solution. Dividing by $h(y)$ assumes $h(y) \ne 0$ so check these separately.
Exponential growth and decay. $y' = ky$ has solutions $y = y(0)e^{kx}$ From $\ln\lvert y\rvert = kx + c$ you get $y = Ae^{kx}$ with $A = \pm e^c$ and $A = 0$ is the constant solution, so $A$ can be any real number.
Initial condition. Use $y(x_0) = y_0$ to find $C$ Then check that the answer satisfies the equation.
Interval of validity. A solution lives on the interval containing $x_0$ where the formula is defined and differentiable. If the formula blows up, the solution ends there; the formula's values on the far side belong to a different solution.
Newton's law of cooling. $T' = -k(T - T_a)$ for room temperature $T_a$ The quantity that decays exponentially is $T - T_a$ not $T$

Builds on: Integration: substitution and parts, Partial fractions and integrating rational functions

Problems

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Solve $\dfrac{dy}{dx} = 3y$ with $y(0) = -2$
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Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 3$
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Solve $\dfrac{dy}{dx} = xy^2$ with $y(0) = 1$ and find the interval on which the solution exists.
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Solve $\dfrac{dy}{dx} = e^{x - y}$ with $y(0) = \ln2$
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A radioactive sample decays at a rate proportional to its mass, $\dfrac{dm}{dt} = -km$ with a half-life of $5$ years. How long until $10\%$ of it is left?
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Coffee at 90°C is left in a 20°C room, and $T' = -k(T - 20)$ After $10$ minutes it is at 60°C. When does it reach 30°C?
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Solve $\dfrac{dy}{dx} = y(1 - y)$ with $y(0) = \tfrac12$
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Gradient flow is gradient descent with an infinitely small step: $\dfrac{dw}{dt} = -\eta L'(w)$ For $L(w) = \tfrac12(w - 3)^2$ $\eta = 0.1$ and $w(0) = 0$ find $w(t)$ and the time at which $w = 2.9$
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Solve $\dfrac{dy}{dx} = 2x(y - 1)$ first with $y(0) = 3$ then with $y(0) = 1$
···
Water drains from a tank through a hole at the bottom, and its depth $h$ (in meters, with $t$ in minutes) satisfies $\dfrac{dh}{dt} = -\tfrac12\sqrt h$ with $h(0) = 16$ When is the tank empty, and when is it half full?

Answers

$y = -2e^{3x}$
$y = \sqrt{x^2 + 9}$ for all $x$
$y = \dfrac{2}{2 - x^2}$ on $-\sqrt2 < x < \sqrt2$
$y = \ln(1 + e^x)$ the softplus function
$\dfrac{5\ln10}{\ln2} \approx 16.61$ years
About $34.77$ minutes
$y = \dfrac{1}{1 + e^{-x}}$ the sigmoid
$w(t) = 3 - 3e^{-0.1t}$ $w = 2.9$ at $t = 10\ln30 \approx 34.01$
$y = 1 + 2e^{x^2}$ for $y(0) = 3$ $y \equiv 1$ for $y(0) = 1$
Empty at $t = 16$ minutes; half full at $t = 16 - 8\sqrt2 \approx 4.69$ minutes

Worked solutions

Problem 1

Solve $\dfrac{dy}{dx} = 3y$ with $y(0) = -2$

$\dfrac{dy}{y} = 3\,dx$ Separate, for $y \ne 0$
$\ln\lvert y\rvert = 3x + c$ Integrate both sides; one constant.
$\lvert y\rvert = e^ce^{3x}$ so $y = Ae^{3x}$ with $A = \pm e^c$ The absolute value is why $A$ can be negative.
$y(0) = A = -2$ The initial condition.
$y = -2e^{3x}$ Check: $y' = -6e^{3x} = 3y$ and $y(0) = -2$ The solution stays negative and grows in size.

Problem 2

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 3$

$y\,dy = x\,dx$ Multiply both sides by $y\,dx$
$\tfrac{y^2}2 = \tfrac{x^2}2 + C$ Integrate both sides.
$C = \tfrac92$ so $y^2 = x^2 + 9$ At $x = 0$ $y = 3$
$y = \sqrt{x^2 + 9}$ Take the positive root, because $y(0) = 3 > 0$
$y = \sqrt{x^2 + 9}$ for all $x$ Check: $y' = \tfrac{x}{\sqrt{x^2 + 9}} = \tfrac xy$ The curve is the upper branch of the hyperbola $y^2 - x^2 = 9$

Problem 3

Solve $\dfrac{dy}{dx} = xy^2$ with $y(0) = 1$ and find the interval on which the solution exists.

$y^{-2}\,dy = x\,dx$ Separate, for $y \ne 0$
$-\dfrac1y = \dfrac{x^2}2 + C$ $\int y^{-2}\,dy = -y^{-1}$
$C = -1$ At $x = 0$ $-1 = 0 + C$
$\dfrac1y = 1 - \dfrac{x^2}2 = \dfrac{2 - x^2}{2}$ so $y = \dfrac{2}{2 - x^2}$ Solve for $y$ only after $C$ is in.
$y = \dfrac{2}{2 - x^2}$ on $-\sqrt2 < x < \sqrt2$ The formula blows up at $x = \pm\sqrt2 \approx \pm1.414$ so the solution through $y(0) = 1$ exists only between them. Growth at a rate $y^2$ is faster than exponential and reaches infinity in finite time.

Problem 4

Solve $\dfrac{dy}{dx} = e^{x - y}$ with $y(0) = \ln2$

$\dfrac{dy}{dx} = e^xe^{-y}$ so $e^y\,dy = e^x\,dx$ Split the exponential, then separate.
$e^y = e^x + C$ Integrate both sides.
$C = 1$ At $x = 0$ $e^{\ln2} = 2 = 1 + C$
$y = \ln(1 + e^x)$ Take the log; $1 + e^x > 0$ so this is defined for every $x$
$y = \ln(1 + e^x)$ the softplus functionCheck: $y' = \tfrac{e^x}{1 + e^x}$ and $e^{x - y} = \tfrac{e^x}{1 + e^x}$ The derivative of softplus is the sigmoid.

Problem 5

A radioactive sample decays at a rate proportional to its mass, $\dfrac{dm}{dt} = -km$ with a half-life of $5$ years. How long until $10\%$ of it is left?

$m = m_0e^{-kt}$ The solution of $m' = -km$ as in Problem 1.
$e^{-5k} = \tfrac12$ so $k = \dfrac{\ln2}{5}$ After one half-life, half the mass is left.
$e^{-kt} = \tfrac1{10}$ so $t = \dfrac{\ln10}{k} = \dfrac{5\ln10}{\ln2}$ Solve for the time when $m = 0.1m_0$
$\dfrac{5\ln10}{\ln2} \approx 16.61$ yearsAbout $3.32$ half-lives: $2^{-3.32} \approx 0.1$ The initial mass cancels, so it was never needed.

Problem 6

Coffee at 90°C is left in a 20°C room, and $T' = -k(T - 20)$ After $10$ minutes it is at 60°C. When does it reach 30°C?

$\dfrac{dT}{T - 20} = -k\,dt$ so $T - 20 = Ae^{-kt}$ Separate and integrate; $T - 20$ decays exponentially.
$A = 90 - 20 = 70$ At $t = 0$
$70e^{-10k} = 40$ so $k = \dfrac{\ln(7/4)}{10} \approx 0.0560$ per minuteAt $t = 10$ $T - 20 = 40$
$70e^{-kt} = 10$ so $t = \dfrac{\ln7}{k} = \dfrac{10\ln7}{\ln(7/4)}$ At $T = 30$ $T - 20 = 10$
About $34.77$ minutes $T$ approaches 20°C but never reaches it.

Problem 7

Solve $\dfrac{dy}{dx} = y(1 - y)$ with $y(0) = \tfrac12$

$\dfrac{dy}{y(1 - y)} = dx$ and $\dfrac{1}{y(1 - y)} = \dfrac1y + \dfrac{1}{1 - y}$ Separate, then partial fractions by the cover-up rule.
$\ln\lvert y\rvert - \ln\lvert1 - y\rvert = x + c$ $\int\tfrac{dy}{1 - y} = -\ln\lvert1 - y\rvert$
$\dfrac{y}{1 - y} = Ae^x$ and $A = 1$ At $x = 0$ $\tfrac{1/2}{1/2} = 1$
$y = e^x(1 - y)$ so $y(1 + e^x) = e^x$ and $y = \dfrac{e^x}{1 + e^x}$ Solve for $y$
$y = \dfrac{1}{1 + e^{-x}}$ the sigmoidDivide top and bottom by $e^x$ This is why $\sigma' = \sigma(1 - \sigma)$ and why logistic growth with rate $1$ and capacity $1$ is a sigmoid. The constant solutions $y \equiv 0$ and $y \equiv 1$ are its two horizontal asymptotes.

Problem 8

Gradient flow is gradient descent with an infinitely small step: $\dfrac{dw}{dt} = -\eta L'(w)$ For $L(w) = \tfrac12(w - 3)^2$ $\eta = 0.1$ and $w(0) = 0$ find $w(t)$ and the time at which $w = 2.9$

$\dfrac{dw}{dt} = -0.1(w - 3)$ $L'(w) = w - 3$
$w - 3 = Ae^{-0.1t}$ and $A = -3$ The same structure as Problem 6: $w - 3$ decays exponentially; $w(0) - 3 = -3$
$w = 3 - 3e^{-0.1t}$ $w$ climbs toward the minimizer $3$
$3e^{-0.1t} = 0.1$ so $t = 10\ln30$ $w = 2.9$ means the gap is $0.1$
$w(t) = 3 - 3e^{-0.1t}$ $w = 2.9$ at $t = 10\ln30 \approx 34.01$ On a quadratic loss the distance to the minimum shrinks by the same factor in every unit of time. Gradient descent with step $\eta$ shrinks it by $1 - \eta$ per step, and $(1 - \eta)^n \approx e^{-\eta n}$

Problem 9

Solve $\dfrac{dy}{dx} = 2x(y - 1)$ first with $y(0) = 3$ then with $y(0) = 1$

$\dfrac{dy}{y - 1} = 2x\,dx$ for $y \ne 1$ Separate.
$\ln\lvert y - 1\rvert = x^2 + c$ so $y = 1 + Ae^{x^2}$ $A = \pm e^c$ is not $0$ so far.
$y(0) = 3$ gives $A = 2$ $1 + A = 3$
For $y(0) = 1$ $y \equiv 1$ satisfies $y' = 0 = 2x(1 - 1)$ The constant solution, which step 1 excluded by dividing by $y - 1$ It is the case $A = 0$
$y = 1 + 2e^{x^2}$ for $y(0) = 3$ $y \equiv 1$ for $y(0) = 1$ Check: $y' = 4xe^{x^2} = 2x\cdot2e^{x^2} = 2x(y - 1)$

Problem 10

Water drains from a tank through a hole at the bottom, and its depth $h$ (in meters, with $t$ in minutes) satisfies $\dfrac{dh}{dt} = -\tfrac12\sqrt h$ with $h(0) = 16$ When is the tank empty, and when is it half full?

$h^{-1/2}\,dh = -\tfrac12\,dt$ Separate, for $h > 0$
$2\sqrt h = -\tfrac t2 + C$ and $C = 8$ $\int h^{-1/2}\,dh = 2h^{1/2}$ at $t = 0$ $2\cdot4 = 8$
$\sqrt h = 4 - \tfrac t4$ so $h = \big(4 - \tfrac t4\big)^2$ for $0 \le t \le 16$ Valid while $\sqrt h \ge 0$
$h = 0$ at $t = 16$ After that $h \equiv 0$ the formula $\big(4 - \tfrac t4\big)^2$ would start rising again, which an empty tank does not do.
$\sqrt8 = 4 - \tfrac t4$ so $t = 16 - 4\sqrt8 = 16 - 8\sqrt2$ Half full means $h = 8$ if the tank has straight sides.
Empty at $t = 16$ minutes; half full at $t = 16 - 8\sqrt2 \approx 4.69$ minutesThe first half drains in under a third of the time, because the flow is fastest when the water is deepest.

Where this goes wrong

1. Adding the constant after exponentiating

The $+C$ is often left until the end, as it would be for an ordinary antiderivative.

$\dfrac{dy}{y} = 3\,dx$ so $\ln\lvert y\rvert = 3x$ Right so far, apart from the constant: Problem 1, step 2.
“Solve for $y$ then add the constant.”The habit that causes the mistake: the constant belongs to the integration step, not to the final line.
$y = e^{3x} + C$ and $y(0) = -2$ gives $y = e^{3x} - 3$ Check it: $y' = 3e^{3x}$ but $3y = 3e^{3x} - 9$ so it does not satisfy the equation. Added at the integration step, $\ln\lvert y\rvert = 3x + c$ becomes $y = Ae^{3x}$ a multiplicative constant, and $y = -2e^{3x}$

2. Splitting a square root

After solving for $y^2$ taking the square root looks like taking it term by term.

$y^2 = x^2 + 9$ Right so far: Problem 2, step 3.
“The square root of $x^2$ is $x$ and of $9$ is $3$ ”The habit that causes the mistake: $\sqrt{a + b}$ treated as $\sqrt a + \sqrt b$
$y = x + 3$ At $x = 4$ this gives $7$ but $\sqrt{16 + 9} = 5$ Check the equation too: $y' = 1$ while $\tfrac xy = \tfrac{x}{x + 3}$ so $y = x + 3$ is not a solution. The answer is $y = \sqrt{x^2 + 9}$

3. Losing the constant solution

Separating means dividing by the function of $y$ and that function can be $0$

$\dfrac{dy}{y - 1} = 2x\,dx$ so $y = 1 + Ae^{x^2}$ with $A = \pm e^c$ Right so far: Problem 9, step 2.
“ $e^c$ is never $0$ so $A \ne 0$ ”The habit that causes the mistake: only the solutions the division allowed are kept.
With $y(0) = 1$ $1 + A = 1$ needs $A = 0$ so there is no solution $y \equiv 1$ is a solution: both sides of $y' = 2x(y - 1)$ are $0$ Dividing by $y - 1$ excluded it. Whenever you divide by $h(y)$ check the values with $h(y) = 0$ separately; here they make $A = 0$ allowed.

4. Forgetting the room temperature in Newton's law

Exponential decay is usually $y = y_0e^{-kt}$ and that pattern gets applied to $T$ itself.

$T' = -k(T - 20)$ $T(0) = 90$ $T(10) = 60$ Right so far: Problem 6.
“ $T$ decays exponentially, so $T = 90e^{-kt}$ ”The habit that causes the mistake: the shift by the room temperature is dropped.
$90e^{-10k} = 60$ gives $k = \tfrac{\ln1.5}{10}$ and $T = 30$ at $t = \tfrac{10\ln3}{\ln1.5} \approx 27.10$ minutes $T = 90e^{-kt}$ decays to 0°C, not to the room's 20°C. It is $T - 20$ that decays: $70e^{-10k} = 40$ and the coffee reaches 30°C at about $34.77$ minutes.

5. Using the formula past the blow-up

The formula for the solution is defined on both sides of the point where it blows up.

$y = \dfrac{2}{2 - x^2}$ solves $y' = xy^2$ with $y(0) = 1$ Right so far: Problem 3, step 4.
“Plug in $x = 2$ ”The habit that causes the mistake: the formula treated as valid wherever it can be evaluated.
$y(2) = \dfrac{2}{2 - 4} = -1$ The solution through $y(0) = 1$ goes to $+\infty$ at $x = \sqrt2 \approx 1.414$ and stops there; a solution must be differentiable on an interval, and it cannot jump from $+\infty$ to negative values. The branch beyond $\sqrt2$ is a different solution, one that never passes through $(0, 1)$ $y(2)$ is not defined for this problem.

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