Chapter 9 · Differential Equations

Exercise 9.3 · Q4: General solution of $\sec^2x\tan y\,dx + \sec^2y\tan x\,dy = 0$

Exercise 9.3Variables separable · Section 9.4.1

The question

For each of the differential equations in Exercises 1 to 10, find the general solution:

4. $\sec^2x\tan y\,dx + \sec^2y\tan x\,dy = 0$

Exam answer

Dividing by $\tan x\tan y$ ( $\ne 0$ ⁠),

\frac{\sec^2x}{\tan x}\,dx + \frac{\sec^2y}{\tan y}\,dy = 0

Integrating both sides,

\begin{aligned} &\int\frac{\sec^2x}{\tan x}\,dx \\ &\quad + \int\frac{\sec^2y}{\tan y}\,dy = \log|C| \end{aligned}

Put $\tan x = t$ ⁠, $\tan y = u$ ⁠. Then $\sec^2x\,dx = dt$ and $\sec^2y\,dy = du$ ⁠.

\begin{aligned} \int\frac{dt}{t} + \int\frac{du}{u} &= \log|C| \\ \log|t| + \log|u| &= \log|C| \\ \log|\tan x\tan y| &= \log|C| \end{aligned}

$\therefore$ $\tan x\tan y = C$ is the general solution.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Each term mixes $x$ and $y$ ⁠, but as a product: $\sec^2x\cdot\tan y$ and $\sec^2y\cdot\tan x$ ⁠. Dividing by $\tan x\tan y$ leaves $\frac{\sec^2x}{\tan x}$ with $dx$ and $\frac{\sec^2y}{\tan y}$ with $dy$ ⁠. Each is a derivative over its function, so it integrates to a logarithm, and writing the constant as $\log|C|$ lets the logs combine.

Given, to find, and the plan

Given: $\sec^2x\tan y\,dx + \sec^2y\tan x\,dy = 0$ ⁠, for $x$ and $y$ where $\tan x$ and $\tan y$ are defined.
To find: The general solution, with one arbitrary constant.
Plan: (1) Divide by $\tan x\tan y$ ⁠, saying where that is allowed. (2) Integrate each term by a substitution, one constant written as $\log|C|$ ⁠. (3) Combine the logs and remove them. (4) Check by implicit differentiation.

Basics you need

1Variables separable

An equation $f_1(x)\,g_1(y)\,dx + f_2(x)\,g_2(y)\,dy = 0$ separates on dividing by $g_1(y)\,f_2(x)$ ⁠, where that is nonzero (Section 9.4.1). Lesson: separate, then integrate

2Derivative over function

With $t = f(x)$ ⁠, $dt = f'(x)\,dx$ ⁠: $\int\frac{f'(x)}{f(x)}\,dx = \int\frac{dt}{t} = \log|f(x)| + C$ (Sections 7.2 and 7.3.1). Basics: every method ends with an integral

3Laws of logarithms

$\log|a| + \log|b| = \log|ab|$ ⁠, and $\log|p| = \log|q|$ gives $|p| = |q|$ ⁠.

4One arbitrary constant

Any form of one constant will do: $C$ ⁠, or $\log|C|$ ⁠, which can also be any real number. Basics: order n gives n constants

Solution

Divide by $\tan x\tan y$ ⁠: the variables separate

Take $x$ and $y$ where $\tan x$ and $\tan y$ are defined ( $\cos x \ne 0$ ⁠, $\cos y \ne 0$ ⁠, so $\sec$ is defined too) and nonzero. Dividing every term by $\tan x\tan y$ ⁠:
$\frac{\sec^2x}{\tan x}\,dx + \frac{\sec^2y}{\tan y}\,dy = 0.$
In the first term $\tan y$ cancels, in the second $\tan x$ cancels, so each term keeps one variable only. Dividing needs $\tan x\tan y \ne 0$ ⁠. The lines where $\tan y = 0$ ( $y = n\pi$ ⁠) do satisfy the equation ( $\tan y = 0$ and $dy = 0$ ⁠); they will come back as $C = 0$ in Step 3.

Section 9.4.1: separate by dividing by a nonzero factor. Lesson: separate, then integrate
Integrate each term, with one constant $\log|C|$

Put $t = \tan x$ ⁠, so $dt = \sec^2x\,dx$ ⁠, and $u = \tan y$ ⁠, so $du = \sec^2y\,dy$ ⁠:
$\int\frac{dt}{t} + \int\frac{du}{u} = \log|C|$ $\log|\tan x| + \log|\tan y| = \log|C|.$
Check each integral by differentiating: $\frac{d}{dx}\log|\tan x| = \frac{\sec^2x}{\tan x}$ ⁠, and the same with $y$ ⁠.

Each fraction is the derivative of $\tan$ over $\tan$ ⁠, so it integrates to $\log|\tan|$ ⁠. The integral of 0 is a constant; we write it as $\log|C|$ with $C \ne 0$ ⁠, which is still one arbitrary constant (as $C$ varies, $\log|C|$ takes every real value). This choice lets the logs combine in the next step.

Sections 7.2 (formula (iv), (xii)) and 7.3.1. Basics: every method ends with an integral
Combine the logs and remove them
$\begin{aligned} \log|\tan x\tan y| &= \log|C| \\ \Rightarrow\ |\tan x\tan y| &= |C|. \end{aligned}$
So $\tan x\tan y = \pm C$ ⁠. Since $C$ is arbitrary, write $C$ for $\pm C$ ⁠:
$\tan x\tan y = C.$
$\tan x\tan y$ is continuous along a solution curve and never 0 in the region of Step 1, so its sign does not change and the $\pm$ is one fixed sign. Allowing $C = 0$ as well adds back the lines $\tan y = 0$ set aside in Step 1. (It also gives the lines $x = n\pi$ ⁠, where $\tan x = 0$ and $dx = 0$ ⁠; they too satisfy the equation as written with $dx$ and $dy$ ⁠.)

$\log|a| + \log|b| = \log|ab|$ ⁠; $\log$ is one-one. Basics: one arbitrary constant
Check: differentiate the answer back

Differentiate $\tan x\tan y = C$ with respect to $x$ ⁠, with $y$ a function of $x$ (product rule and chain rule):
$\sec^2x\tan y + \tan x\sec^2y\,\frac{dy}{dx} = 0.$
Multiplying by $dx$ gives $\sec^2x\tan y\,dx + \sec^2y\tan x\,dy = 0$ ⁠, the given equation.

The constant has disappeared, so the given equation holds on every member of the family. This is the substitution check for an answer left in implicit form.

Product rule and chain rule (Chapter 5). Basics: y′ is the slope

Result

The general solution is $\tan x\tan y = C$ ⁠, where $C$ is an arbitrary constant, for $x$ and $y$ with $\cos x \ne 0$ and $\cos y \ne 0$ ⁠. (Textbook answer: $\tan x\tan y = C$ ⁠.)

Check it with numbers

Take $C = 2$ and $0 < x < \frac{\pi}{2}$ ⁠: then $\tan y = \frac{2}{\tan x}$ ⁠, so $y = \tan^{-1}\frac{2}{\tan x}$ ⁠. At $x = 0.5$ ⁠, $y \approx 1.3041496999$ ⁠, and the equation $\frac{dy}{dx} = -\frac{\sec^2x\tan y}{\sec^2y\tan x}$ gives the slope $\approx -0.60414684$ ⁠.

The curve itself: $y(0.501) \approx 1.3035453225$ and $y(0.499) \approx 1.3047536166$ ⁠, so

\begin{aligned} &\frac{1.3035453225 - 1.3047536166}{0.002} \\ &\approx -0.60415. \end{aligned}

Common mistakes

Dividing by $\sec^2x\sec^2y$ instead. That leaves $\tan y\,dx$ and $\tan x\,dy$ ⁠, still mixed.
Getting $\tan x + \tan y = C$ from $\log|\tan x| + \log|\tan y|$ ⁠. A sum of logs is the log of a product.
A constant on each side. Integrate the whole equation once and add one constant.

What to take away

For $f_1(x)g_1(y)\,dx + f_2(x)g_2(y)\,dy = 0$ ⁠, divide by $g_1(y)f_2(x)$ ⁠.
$\int\frac{f'(x)}{f(x)}\,dx = \log|f(x)|$ ⁠: spot "derivative over function".
When every term is a log, write the constant as $\log|C|$ and combine.

Every fact used, and where it comes from

Fact	Used in	Source
Variables separable: divide by a nonzero factor and integrate	Step 1	Section 9.4.1 (Lesson)
$\frac{d}{dx}\tan x = \sec^2x$ ⁠; $\int\frac{dt}{t} = \log\|t\| + C$	Step 2	Section 7.2, formulas (iv), (xii) (Basics)
Integration by substitution	Step 2	Section 7.3.1
One arbitrary constant, here written $\log\|C\|$	Steps 2, 3	Section 9.3 (Basics)
$\log\|a\| + \log\|b\| = \log\|ab\|$	Step 3	Laws of logarithms
Product rule and chain rule	Step 4	Chapter 5 (Basics)

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Exercise 9.3 · Q4: General solution of sec⁡2xtan⁡y dx+sec⁡2ytan⁡x dy=0\sec^2x\tan y\,dx + \sec^2y\tan x\,dy = 0sec2xtanydx+sec2ytanxdy=0

Basics you need

Solution

Divide by tan⁡xtan⁡y\tan x\tan ytanxtany⁠: the variables separate

Integrate each term, with one constant log⁡∣C∣\log|C|log∣C∣

Combine the logs and remove them

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 9.3 · Q4: General solution of $\sec^2x\tan y\,dx + \sec^2y\tan x\,dy = 0$

Divide by $\tan x\tan y$ ⁠: the variables separate

Integrate each term, with one constant $\log|C|$