Chapter 9 · Differential Equations

Exercise 9.3 · Q13: Particular solution of $\cos\left(\frac{dy}{dx}\right) = a$ ⁠, $y(0) = 1$

Exercise 9.3Variables separable, particular solution · Sections 9.3 and 9.4.1

The question

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:

13. $\cos\left(\dfrac{dy}{dx}\right) = a$ $(a \in \mathbf{R})$ ⁠; $y = 1$ when $x = 0$

Exam answer

Given $\cos\left(\frac{dy}{dx}\right) = a$ ⁠, where $-1 \le a \le 1$ ⁠.

\frac{dy}{dx} = \cos^{-1}a

Separating the variables, $dy = \cos^{-1}a\,dx$ ⁠.

Integrating both sides,

y = x\cos^{-1}a + C\quad\ldots(\text{i})

Putting $x = 0$ ⁠, $y = 1$ in (i): $1 = 0 + C$ ⁠, so $C = 1$ ⁠.

\begin{aligned} y &= x\cos^{-1}a + 1 \\ \frac{y - 1}{x} &= \cos^{-1}a\quad(x \ne 0) \end{aligned}

$\therefore$ $\cos\left(\dfrac{y - 1}{x}\right) = a$ is the particular solution.

The answer key prints $\cos\left(\frac{y - 2}{x}\right) = a$ ⁠; the condition $y = 1$ when $x = 0$ gives $y - 1$ ⁠, not $y - 2$ ⁠.

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

The one idea this problem is built on

The equation says the cosine of the slope is the fixed number $a$ ⁠, so the slope itself is a constant, $\cos^{-1}a$ ⁠. A curve whose slope never changes is a straight line, and the condition $y = 1$ at $x = 0$ fixes where it crosses the $y$ -axis. Dividing by $x$ and taking the cosine turns the line into the form $\cos\left(\frac{y - 1}{x}\right) = a$ ⁠.

Given, to find, and the plan

Given: $\cos\left(\frac{dy}{dx}\right) = a$ with $a$ a real constant, and $y = 1$ when $x = 0$ ⁠.
To find: The particular solution: no arbitrary constant, passing through $(0, 1)$ ⁠.
Plan: (1) Order and degree, and which $a$ are possible. (2) Solve for the slope. (3) Separate and integrate, one constant $C$ ⁠. (4) Use the condition and write the answer in the cosine form. (5) Check by substituting back. (6) Look at the other possible slopes and at the answer key's form.

Basics you need

1Order and degree

The order is the order of the highest derivative. The degree is defined only when the equation is a polynomial equation in the derivatives (Section 9.2.2). Lesson: order and degree

2Inverse cosine

$\cos^{-1}a$ is defined for $-1 \le a \le 1$ ⁠, its principal value lies in $[0, \pi]$ ⁠, and $\cos(\cos^{-1}a) = a$ (Chapter 2).

3Variables 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 $g$ is a constant and $h(y) = 1$ ⁠. Lesson: separate, then integrate

4Particular solution

Put the given $x$ and $y$ into the general solution and solve for the constant (Section 9.3). Basics: one constant, one condition

Solution

Order, degree, and which values of $a$ are possible

The only derivative is $\frac{dy}{dx}$ ⁠, so the order is 1. The equation is not a polynomial equation in $\frac{dy}{dx}$ (the derivative sits inside a cosine), so its degree is not defined.

The cosine of a real number always lies between $-1$ and $1$ ⁠. So a real slope with $\cos\left(\frac{dy}{dx}\right) = a$ exists only when $-1 \le a \le 1$ ⁠.

The book writes $a \in \mathbf{R}$ ⁠, but for $a = 2$ ⁠, say, no real number has cosine 2, and the equation has no solution at all. So we work with $|a| \le 1$ throughout. One constant will appear, because the order is 1.

Section 9.2.2: degree is defined only for polynomial equations in the derivatives. Lesson: order and degree
Solve for the slope

Take $\cos^{-1}$ of both sides:
$\frac{dy}{dx} = \cos^{-1}a.$
$\cos^{-1}a$ is the angle in $[0, \pi]$ whose cosine is $a$ ⁠, so this slope does satisfy $\cos\left(\frac{dy}{dx}\right) = a$ ⁠. It is a fixed number: it does not depend on $x$ or $y$ ⁠. Other angles also have cosine $a$ (for example $-\cos^{-1}a$ ⁠); this is the choice the book's method makes, and Step 6 shows that the final cosine form covers the others too.

Principal value of $\cos^{-1}$ (Chapter 2). Basics: y′ is the slope
Separate the variables and integrate

Multiply by $dx$ and integrate both sides:
$\begin{aligned} dy &= \cos^{-1}a\,dx \\ \int dy &= \cos^{-1}a\int dx \\ y &= x\cos^{-1}a + C. \end{aligned}$
This is the separable form with $g(x) = \cos^{-1}a$ (a constant) and $h(y) = 1$ ⁠, so nothing in $y$ is divided by. A constant comes out of the integral, and $\int dx = x$ ⁠. The constants of both sides combine into one $C$ ⁠: this is the general solution, the family of all lines with slope $\cos^{-1}a$ ⁠.

Section 9.4.1: separate, then integrate. Lesson: separate, then integrate
Use the condition, then write the cosine form

Put $x = 0$ ⁠, $y = 1$ ⁠: $1 = 0\cdot\cos^{-1}a + C$ ⁠, so $C = 1$ and
$y = x\cos^{-1}a + 1.$
For $x \ne 0$ ⁠, subtract 1 and divide by $x$ ⁠, then take the cosine of both sides:
$\frac{y - 1}{x} = \cos^{-1}a \;\Rightarrow\; \cos\left(\frac{y - 1}{x}\right) = a.$
One condition fixes the one constant. The last step uses $\cos(\cos^{-1}a) = a$ ⁠, which holds because $|a| \le 1$ ⁠. The cosine form is the textbook's style of answer; it needs $x \ne 0$ because of the division, while the line $y = x\cos^{-1}a + 1$ itself passes through $(0, 1)$ ⁠.

Section 9.3: a particular solution gives the constant a value. Basics: one constant, one condition
Check: substitute back

From $y = x\cos^{-1}a + 1$ ⁠: $\frac{dy}{dx} = \cos^{-1}a$ ⁠, so $\cos\left(\frac{dy}{dx}\right) = \cos(\cos^{-1}a) = a$ ⁠. At $x = 0$ ⁠: $y = 0 + 1 = 1$ ⁠.

Both the equation and the condition hold, for every $a$ with $-1 \le a \le 1$ ⁠. Substituting back is the definition of a solution (Section 9.3).

A solution makes L.H.S. = R.H.S. Lesson: general and particular solutions
The other slopes, and the answer key's form

Any line $y = bx + 1$ whose slope $b$ has $\cos b = a$ also satisfies the equation and passes through $(0, 1)$ ⁠; for $a = \frac{1}{2}$ the slopes $b = \frac{\pi}{3}$ and $b = -\frac{\pi}{3}$ both work. For every such line $\frac{y - 1}{x} = b$ ⁠, so each of them satisfies the same relation $\cos\left(\frac{y - 1}{x}\right) = a$ ⁠. The cosine form describes them all at once.

The answer key prints $\cos\left(\frac{y - 2}{x}\right) = a$ ⁠. That relation describes lines $y = bx + 2$ ⁠, which pass through $(0, 2)$ ⁠, not $(0, 1)$ ⁠. The condition $y = 1$ when $x = 0$ gives $y - 1$ ⁠, so the key's $y - 2$ is a misprint.

Because the equation is not of first degree in $\frac{dy}{dx}$ ⁠, the given point does not pick out a single slope; it picks out the intercept 1. The relation $\cos\left(\frac{y - 1}{x}\right) = a$ is exactly "intercept 1 and a slope whose cosine is $a$ ". With the key's $y - 2$ ⁠, putting $x = 0$ would need $y = 2$ ⁠.

A particular solution must pass through the given point. Lesson: general and particular solutions

Result

For $-1 \le a \le 1$ ⁠, the particular solution is $y = x\cos^{-1}a + 1$ ⁠, that is, $\cos\left(\frac{y - 1}{x}\right) = a$ ( $x \ne 0$ ⁠). For $|a| > 1$ there is no solution, so the printed $a \in \mathbf{R}$ really needs $|a| \le 1$ ⁠. (Textbook answer: $\cos\left(\frac{y - 2}{x}\right) = a$ ⁠, a misprint for $\cos\left(\frac{y - 1}{x}\right) = a$ ⁠: the key's form gives $y = 2$ at $x = 0$ ⁠.)

Check it with numbers

Take $a = 0.5$ ⁠: $\cos^{-1}0.5 = \frac{\pi}{3} \approx 1.0471976$ ⁠, so the line is $y = 1.0471976x + 1$ ⁠.

At $x = 2$ ⁠: $y \approx 3.0943951$ ⁠, and $\frac{y - 1}{x} \approx \frac{2.0943951}{2} = 1.0471976$ ⁠, whose cosine is $0.5 = a$ ⁠.

The key's form at the same point: $\frac{y - 2}{x} \approx 0.5471976$ ⁠, whose cosine is about $0.8540$ ⁠, not $0.5$ ⁠.

Common mistakes

Writing $\frac{dy}{dx} = \cos a$ ⁠. The cosine is applied to the slope, so the slope is $\cos^{-1}a$ ⁠.
Ignoring $|a| \le 1$ ⁠. For $|a| > 1$ the equation has no real solution.
Copying the key's $y - 2$ ⁠. Put $x = 0$ ⁠, $y = 1$ into your answer: only $y - 1$ fits.

What to take away

If an equation forces $\frac{dy}{dx}$ to be a constant, the solutions are straight lines and the condition fixes the intercept.
An equation like $\cos\left(\frac{dy}{dx}\right) = a$ has order 1 but no degree.
Always test the final answer at the given point.

Every fact used, and where it comes from

Fact	Used in	Source
Order; degree only for polynomial equations in the derivatives	Step 1	Sections 9.2.1, 9.2.2 (Lesson)
$\cos^{-1}a$ exists for $\|a\| \le 1$ ⁠, lies in $[0, \pi]$ ⁠; $\cos(\cos^{-1}a) = a$	Steps 1, 2, 4	Chapter 2, Section 2.2
Variables separable: separate, then integrate	Step 3	Section 9.4.1 (Lesson)
General solution has an arbitrary constant; a particular solution gives it a value	Steps 3, 4	Section 9.3 (Basics)
A solution makes L.H.S. = R.H.S. and passes through the given point	Steps 5, 6	Section 9.3 (Lesson)

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Exercise 9.3 · Q13: Particular solution of cos⁡(dydx)=a\cos\left(\frac{dy}{dx}\right) = acos(dxdy​)=a⁠, y(0)=1y(0) = 1y(0)=1

Basics you need

Solution

Order, degree, and which values of aaa are possible

Solve for the slope

Separate the variables and integrate

Use the condition, then write the cosine form

Check: substitute back

The other slopes, and the answer key's form

Every fact used, and where it comes from

Exercise 9.3 · Q13: Particular solution of $\cos\left(\frac{dy}{dx}\right) = a$ ⁠, $y(0) = 1$

Order, degree, and which values of $a$ are possible