Chapter 9 · Differential Equations

Exercise 9.3 · Q10: General solution of $e^x\tan y\,dx + (1 - e^x)\sec^2y\,dy = 0$

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:

10. $e^x\tan y\,dx + (1 - e^x)\sec^2y\,dy = 0$

Exam answer

Dividing by $(1 - e^x)\tan y$ ( $\ne 0$ ⁠),

\frac{e^x}{1 - e^x}\,dx + \frac{\sec^2y}{\tan y}\,dy = 0

Integrating both sides,

\begin{aligned} &\int\frac{e^x}{1 - e^x}\,dx \\ &\quad + \int\frac{\sec^2y}{\tan y}\,dy = \log|C| \end{aligned}

Put $1 - e^x = t$ ⁠, $\tan y = u$ ⁠. Then $-e^x\,dx = dt$ 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 y| &= \log|C| + \log|t| \\ \log|\tan y| &= \log|C(1 - e^x)| \end{aligned}

$\therefore$ $\tan y = C(1 - e^x)$ is the general solution.

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

The one idea this problem is built on

Both terms are products of an $x$ part and a $y$ part: $e^x\cdot\tan y$ and $(1 - e^x)\cdot\sec^2y$ ⁠. Dividing by $(1 - e^x)\tan y$ separates them. Each resulting fraction is "derivative over function" (up to a sign), so both integrate to logarithms, and a constant written as $\log|C|$ combines everything into $\tan y = C(1 - e^x)$ ⁠.

Given, to find, and the plan

Given: $e^x\tan y\,dx + (1 - e^x)\sec^2y\,dy = 0$ ⁠, for $y$ with $\cos y \ne 0$ ⁠.
To find: The general solution, with one arbitrary constant.
Plan: (1) Divide by $(1 - e^x)\tan y$ ⁠, saying where that is allowed. (2) Integrate each term by a substitution, one constant $\log|C|$ ⁠. (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 $t = f(x)$ ⁠: $\int\frac{f'(x)}{f(x)}\,dx = \log|f(x)| + C$ (Sections 7.2 and 7.3.1). Here $\frac{d}{dx}(1 - e^x) = -e^x$ and $\frac{d}{dy}\tan y = \sec^2y$ ⁠. Basics: every method ends with an integral

3Laws of logarithms

$\log|a| + \log|b| = \log|ab|$ ⁠, and equal logs mean equal moduli.

4One arbitrary constant

Writing the constant as $\log|C|$ is still one arbitrary constant. Basics: order n gives n constants

Solution

Divide by $(1 - e^x)\tan y$ ⁠: the variables separate

Take $x \ne 0$ (so $1 - e^x \ne 0$ ⁠) and $y$ with $\cos y \ne 0$ and $\tan y \ne 0$ ⁠. Dividing every term by $(1 - e^x)\tan y$ ⁠:
$\frac{e^x}{1 - e^x}\,dx + \frac{\sec^2y}{\tan y}\,dy = 0.$
In the first term $\tan y$ cancels and in the second $1 - e^x$ cancels, so each term keeps one variable. Dividing needs $1 - e^x \ne 0$ (that is $x \ne 0$ ⁠) and $\tan y \ne 0$ ⁠. The lines $\tan y = 0$ ( $y = n\pi$ ⁠) satisfy the equation ( $\tan y = 0$ ⁠, $dy = 0$ ⁠); they return 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 $t = 1 - e^x$ ⁠, so $dt = -e^x\,dx$ ⁠, and $u = \tan y$ ⁠, so $du = \sec^2y\,dy$ ⁠:
$-\int\frac{dt}{t} + \int\frac{du}{u} = \log|C|$ $-\log|1 - e^x| + \log|\tan y| = \log|C|.$
Check each integral by differentiating: $\frac{d}{dy}\log|\tan y| = \frac{\sec^2y}{\tan y}$ ⁠, and
$\begin{aligned} &\frac{d}{dx}\left(-\log|1 - e^x|\right) \\ &= -\frac{-e^x}{1 - e^x} = \frac{e^x}{1 - e^x}. \end{aligned}$
The top $e^x$ is minus the derivative of $1 - e^x$ ⁠, which explains the minus sign in front of the first log. The integral of 0 is a constant, written $\log|C|$ ( $C \ne 0$ ⁠) so that the logs combine.

Section 7.2, formulas (iv), (xi), (xii), with substitutions (Section 7.3.1). Basics: every method ends with an integral
Combine the logs and remove them
$\begin{aligned} \log|\tan y| &= \log|C| + \log|1 - e^x| \\ &= \log|C(1 - e^x)| \end{aligned}$
so $|\tan y| = |C(1 - e^x)|$ and, writing $C$ for $\pm C$ ⁠,
$\tan y = C(1 - e^x).$
$\log$ is one-one, so the moduli are equal. Along a solution curve $\frac{\tan y}{1 - e^x}$ is continuous and never 0 here, so its sign is fixed and absorbed into $C$ ⁠. Allowing $C = 0$ adds back the lines $\tan y = 0$ ⁠.

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

Differentiate $\tan y = C(1 - e^x)$ with respect to $x$ ⁠:
$\sec^2y\,\frac{dy}{dx} = -Ce^x.$
From the answer, $C = \frac{\tan y}{1 - e^x}$ ⁠, so
$\sec^2y\,\frac{dy}{dx} = -\frac{e^x\tan y}{1 - e^x}.$
Multiplying by $(1 - e^x)\,dx$ gives $e^x\tan y\,dx + (1 - e^x)\sec^2y\,dy = 0$ ⁠, the given equation.

Eliminating $C$ gives back the equation itself, so every member of the family is a solution (Section 9.3).

Chain rule (Chapter 5). Basics: y′ is the slope

Result

The general solution is $\tan y = C(1 - e^x)$ ⁠, where $C$ is an arbitrary constant, for $y$ with $\cos y \ne 0$ ⁠. (Textbook answer: $\tan y = C(1 - e^x)$ ⁠.)

Check it with numbers

Take $C = 1$ ⁠: $y = \tan^{-1}(1 - e^x)$ ⁠. At $x = 0.5$ ⁠, $y \approx -0.5754757643$ ⁠, and the equation $\frac{dy}{dx} = -\frac{e^x\tan y}{(1 - e^x)\sec^2y}$ gives the slope $\approx -1.16038547$ ⁠.

The curve itself: $y(0.501) \approx -0.5766358559$ and $y(0.499) \approx -0.5743150861$ ⁠, so

\begin{aligned} &\frac{-0.5766358559 + 0.5743150861}{0.002} \\ &\approx -1.1604. \end{aligned}

Common mistakes

$\int\frac{e^x}{1 - e^x}\,dx = \log|1 - e^x|$ ⁠. The derivative of $1 - e^x$ is $-e^x$ ⁠, so the integral is $-\log|1 - e^x|$ ⁠.
Ending with $\tan y = \frac{C}{1 - e^x}$ after a sign slip. Check by differentiating back.
Forgetting that $x \ne 0$ was needed to divide by $1 - e^x$ ⁠.

What to take away

Divide by the "wrong-variable" factors, here $(1 - e^x)\tan y$ ⁠, to separate.
Watch the sign when the top is minus the derivative of the bottom.
Log on every term: 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)
$\frac{d}{dx}e^x = e^x$ ⁠; $\frac{d}{dy}\tan y = \sec^2y$ ⁠; $\int\frac{dt}{t} = \log\|t\| + C$	Step 2	Section 7.2, formulas (xi), (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
Chain rule	Step 4	Chapter 5 (Basics)

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Exercise 9.3 · Q10: General solution of extan⁡y dx+(1−ex)sec⁡2y dy=0e^x\tan y\,dx + (1 - e^x)\sec^2y\,dy = 0extanydx+(1−ex)sec2ydy=0

Basics you need

Solution

Divide by (1−ex)tan⁡y(1 - e^x)\tan y(1−ex)tany⁠: 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 · Q10: General solution of $e^x\tan y\,dx + (1 - e^x)\sec^2y\,dy = 0$

Divide by $(1 - e^x)\tan y$ ⁠: the variables separate

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