Chapter 9 · Differential Equations

Exercise 9.3 · Q5: General solution of $(e^x + e^{-x})\,dy - (e^x - e^{-x})\,dx = 0$

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:

5. $(e^x + e^{-x})\,dy - (e^x - e^{-x})\,dx = 0$

Exam answer

Since $e^x + e^{-x} > 0$ ⁠, separating the variables,

dy = \frac{e^x - e^{-x}}{e^x + e^{-x}}\,dx

Integrating both sides,

y = \int\frac{e^x - e^{-x}}{e^x + e^{-x}}\,dx

Put $e^x + e^{-x} = t$ ⁠. Then $(e^x - e^{-x})\,dx = dt$ ⁠.

\begin{aligned} y &= \int\frac{dt}{t} = \log|t| + C \\ &= \log(e^x + e^{-x}) + C \end{aligned}

$\therefore$ $y = \log(e^x + e^{-x}) + C$ is the general solution.

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

The one idea this problem is built on

The $dy$ term carries only $x$ as a coefficient, so dividing by $e^x + e^{-x}$ ⁠, which is always positive, leaves $dy$ on one side and a function of $x$ on the other. That function is "derivative over function": the top $e^x - e^{-x}$ is the derivative of the bottom $e^x + e^{-x}$ ⁠, so it integrates to a logarithm.

Given, to find, and the plan

Given: $(e^x + e^{-x})\,dy - (e^x - e^{-x})\,dx = 0$ ⁠, for all real $x$ ⁠.
To find: The general solution, with one arbitrary constant.
Plan: (1) Separate by dividing by $e^x + e^{-x} > 0$ ⁠. (2) Integrate by the substitution $t = e^x + e^{-x}$ ⁠, one constant. (3) Check by differentiating back.

Basics you need

1Variables separable

If $\frac{dy}{dx} = g(x)\,h(y)$ with $h(y) \ne 0$ ⁠, then $\int\frac{dy}{h(y)} = \int g(x)\,dx$ (Section 9.4.1). Here $h(y) = 1$ ⁠. Lesson: separate, then integrate

2Derivatives of $e^x$ and $e^{-x}$

$\frac{d}{dx}e^x = e^x$ and $\frac{d}{dx}e^{-x} = -e^{-x}$ (chain rule). Basics: the exponential function

3Derivative over function

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

4One arbitrary constant

A first order equation has one arbitrary constant in its general solution. Basics: order n gives n constants

Solution

Separate: divide by $e^x + e^{-x}$

Move the $dx$ term across: $(e^x + e^{-x})\,dy = (e^x - e^{-x})\,dx$ ⁠. Both $e^x$ and $e^{-x}$ are positive, so $e^x + e^{-x} > 0$ ⁠, and we may divide by it:
$dy = \frac{e^x - e^{-x}}{e^x + e^{-x}}\,dx.$
The left side is now $dy$ alone and the right side holds only $x$ ⁠. In the book's form, $\frac{dy}{dx} = g(x)\cdot 1$ ⁠, so $h(y) = 1$ and no value of $y$ needs to be set aside. The divisor is never 0, so the equation holds for every real $x$ ⁠.

Section 9.4.1: separate, then integrate. Lesson: separate, then integrate
Integrate by the substitution $t = e^x + e^{-x}$

Put $t = e^x + e^{-x}$ ⁠. Then $dt = (e^x - e^{-x})\,dx$ ⁠, exactly the top times $dx$ ⁠:
$\begin{aligned} y &= \int\frac{e^x - e^{-x}}{e^x + e^{-x}}\,dx = \int\frac{dt}{t} \\ &= \log|t| + C = \log(e^x + e^{-x}) + C. \end{aligned}$
Check the integral by differentiating:
$\frac{d}{dx}\log(e^x + e^{-x}) = \frac{e^x - e^{-x}}{e^x + e^{-x}}.$
The derivative of $e^x + e^{-x}$ is $e^x - e^{-x}$ (the chain rule gives $-e^{-x}$ for $e^{-x}$ ⁠), so the fraction is "derivative over function". Since $t = e^x + e^{-x} > 0$ ⁠, $|t| = t$ and no modulus is needed. The constants from $\int dy$ and from the right side combine into one $C$ ⁠.

Section 7.2, formula (xii), with a substitution (Section 7.3.1). Basics: every method ends with an integral
One constant, and the general solution
$y = \log(e^x + e^{-x}) + C.$
This has one arbitrary constant, as a first order equation needs, and gives $y$ explicitly. It is exactly the textbook's form.

Section 9.3: general solution. Basics: order n gives n constants
Check: substitute back

For $y = \log(e^x + e^{-x}) + C$ ⁠, Step 2 gives $\frac{dy}{dx} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$ ⁠, so
$(e^x + e^{-x})\,dy = (e^x - e^{-x})\,dx,$
which is the given equation. It holds for every $C$ and every real $x$ ⁠.

Substituting back is the definition of a solution (Section 9.3). Multiplying by $e^x + e^{-x}$ is safe because it is positive.

A solution makes L.H.S. = R.H.S. Lesson: general and particular solutions

Result

The general solution is $y = \log(e^x + e^{-x}) + C$ ⁠, where $C$ is an arbitrary constant, valid for every real $x$ ⁠. (Textbook answer: $y = \log(e^x + e^{-x}) + C$ ⁠.)

Check it with numbers

Take $C = 0$ ⁠. At $x = 0.5$ ⁠, $y = \log(e^{0.5} + e^{-0.5}) \approx 0.8132616875$ ⁠, and the equation gives the slope $\frac{e^{0.5} - e^{-0.5}}{e^{0.5} + e^{-0.5}}$ ⁠, which is $\approx 0.46211716$ ⁠.

The curve itself: $y(0.501) \approx 0.8137241978$ and $y(0.499) \approx 0.8127999637$ ⁠, so

\begin{aligned} &\frac{0.8137241978 - 0.8127999637}{0.002} \\ &\approx 0.46212. \end{aligned}

Common mistakes

Splitting the fraction into $\frac{e^x}{e^x + e^{-x}} - \frac{e^{-x}}{e^x + e^{-x}}$ ⁠. Neither piece is easier; keep it whole and substitute.
Sign of $\frac{d}{dx}e^{-x}$ ⁠. It is $-e^{-x}$ ⁠; that minus is what makes the top the derivative of the bottom.
Writing $\log(e^x) + \log(e^{-x})$ for $\log(e^x + e^{-x})$ ⁠. The log of a sum is not a sum of logs.

What to take away

Divide by the coefficient of $dy$ when it depends on $x$ only and is never 0.
Spot "derivative over function": $\int\frac{f'(x)}{f(x)}\,dx = \log|f(x)|$ ⁠.
A positive argument needs no modulus: $\log(e^x + e^{-x})$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Variables separable: separate, then integrate	Step 1	Section 9.4.1 (Lesson)
$\frac{d}{dx}e^x = e^x$ ⁠; $\frac{d}{dx}e^{-x} = -e^{-x}$	Step 2	Section 7.2, formula (xi); chain rule (Basics)
$\int\frac{dt}{t} = \log\|t\| + C$ ⁠; substitution	Step 2	Section 7.2, formula (xii); Section 7.3.1 (Basics)
One arbitrary constant for a first order equation	Step 3	Section 9.3 (Basics)
A solution makes L.H.S. = R.H.S.	Step 4	Section 9.3 (Lesson)

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Exercise 9.3 · Q5: General solution of (ex+e−x) dy−(ex−e−x) dx=0(e^x + e^{-x})\,dy - (e^x - e^{-x})\,dx = 0(ex+e−x)dy−(ex−e−x)dx=0

Basics you need

Solution

Separate: divide by ex+e−xe^x + e^{-x}ex+e−x

Integrate by the substitution t=ex+e−xt = e^x + e^{-x}t=ex+e−x

One constant, and the general solution

Check: substitute back

Every fact used, and where it comes from

Exercise 9.3 · Q5: General solution of $(e^x + e^{-x})\,dy - (e^x - e^{-x})\,dx = 0$

Separate: divide by $e^x + e^{-x}$

Integrate by the substitution $t = e^x + e^{-x}$