Chapter 9 · Differential Equations

Exercise 9.2 · Q9: $x + y = \tan^{-1}y$ solves $y^2y' + y^2 + 1 = 0$

Exercise 9.2Verifying an implicit solution · Section 9.3

The question

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

9. $x + y = \tan^{-1}y$ : $y^2y' + y^2 + 1 = 0$

Exam answer

Given $x + y = \tan^{-1}y$ ⁠. Differentiating both sides with respect to $x$ ⁠,

1 + y' = \frac{1}{1 + y^2}\,y'

Multiplying by $1 + y^2$ ⁠,

\begin{aligned} (1 + y^2)(1 + y') &= y' \\ \Rightarrow\ 1 + y' + y^2 + y^2y' &= y' \\ \Rightarrow\ y^2y' + y^2 + 1 &= 0 \end{aligned}

which is the given differential equation.

$\therefore$ $x + y = \tan^{-1}y$ is a solution of the given differential equation. Hence verified.

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

The one idea this problem is built on

The solution is implicit, so differentiate the relation with respect to $x$ ⁠. The derivative of $\tan^{-1}y$ brings the fraction $\frac{1}{1 + y^2}$ ⁠; multiplying through by $1 + y^2$ clears it, and after expanding, the two $y'$ terms on opposite sides cancel. What is left is the given equation.

Given, to find, and the plan

Given: The relation $x + y = \tan^{-1}y$ ⁠, defining $y$ implicitly as a function of $x$ ⁠; the equation $y^2y' + y^2 + 1 = 0$ ⁠.
To show: That the function $y$ defined by the relation satisfies the equation wherever it is differentiable.
Plan: (1) Differentiate implicitly. (2) Clear the fraction. (3) Expand and simplify to the given equation. (4) Find where $y'$ exists, and conclude.

Basics you need

1Derivative of $\tan^{-1}$

$\frac{d}{dy}\tan^{-1}y = \frac{1}{1 + y^2}$ (Section 5.3.3); with $y$ a function of $x$ ⁠, the chain rule adds the factor $y'$ ⁠. Basics: y′ and y″

2Implicit differentiation

Differentiate both sides of the relation with respect to $x$ (Section 5.3.2). Basics: y′ and y″

3What a solution is

A relation whose $y$ satisfies the equation is a solution; Exercise 9.2 allows implicit ones (Section 9.3). Lesson: general and particular solutions

Solution

Differentiate both sides with respect to $x$
$\begin{aligned} \frac{d}{dx}(x + y) &= \frac{d}{dx}\tan^{-1}y \\ \Rightarrow\ 1 + y' &= \frac{y'}{1 + y^2}. \end{aligned}$
On the left, $x$ gives 1 and $y$ gives $y'$ ⁠. On the right, the chain rule gives $\frac{1}{1 + y^2}\cdot y'$ ⁠.

$(\tan^{-1}u)' = \frac{u'}{1 + u^2}$ (Section 5.3.3). Basics: y′ and y″
Clear the fraction
$(1 + y^2)(1 + y') = y'.$
Multiply both sides by $1 + y^2$ ⁠, which is at least 1, never 0, so nothing is lost or gained.

Multiplying an equation by a non-zero quantity gives an equivalent equation. Lesson: is it a solution?
Expand and simplify
$\begin{aligned} 1 + y' + y^2 + y^2y' &= y' \\ y^2y' + y^2 + 1 &= 0. \end{aligned}$
This is the given equation, so L.H.S. = R.H.S.

Expanding the product gives four terms. The $y'$ on the left cancels the $y'$ on the right, and the remaining terms are rearranged in the book's order.

Section 9.3: the function satisfies the equation. Lesson: is it a solution?
Where $y'$ exists, and conclusion

The relation reads $x = \tan^{-1}y - y$ ⁠. As $y$ increases, this decreases (its rate $\frac{1}{1 + y^2} - 1 = -\frac{y^2}{1 + y^2}$ is negative except at $y = 0$ ⁠) and takes every real value, so the relation defines $y$ as a function of $x$ for every real $x$ ⁠. From the equation, $y' = -\frac{1 + y^2}{y^2}$ ⁠, which exists for $y \ne 0$ ⁠. At $y = 0$ (that is $x = 0$ ⁠) the tangent is vertical; there $y^2y' + y^2 + 1$ would equal 1, not 0, so no derivative could satisfy it.

So L.H.S. = R.H.S. at every $x \ne 0$ ⁠: hence verified.

The equation has order 1 and degree 1 (a polynomial equation in $y'$ ⁠, power 1). The relation has no arbitrary constant, so it is a particular solution; $x + y = \tan^{-1}y + C$ would be the general one.

The check needs $y'$ to exist. Locating the one point where it fails ( $y = 0$ ⁠) completes the domain statement.

A solution satisfies the equation where it is defined and differentiable (Section 9.3). Basics: order n gives n constants

Result

Differentiating $x + y = \tan^{-1}y$ gives $1 + y' = \frac{y'}{1 + y^2}$ ⁠, hence $(1 + y^2)(1 + y') = y'$ ⁠, that is $y^2y' + y^2 + 1 = 0$ ⁠. So the relation is a solution of the given equation at every $x \ne 0$ (where $y \ne 0$ and $y'$ exists). (The answer key gives no answer for this question, since it is a verification.)

Check it with numbers

Take $y = 1$ ⁠: then $x = \tan^{-1}1 - 1 = \frac{\pi}{4} - 1 \approx -0.214602$ ⁠. The equation gives $y' = -\frac{1 + 1}{1} = -2$ ⁠, and $y^2y' + y^2 + 1 = -2 + 1 + 1 = 0$ ⁠.

Solving the relation numerically: $y(x + 0.001) \approx 0.9979980$ and $y(x - 0.001) \approx 1.0019980$ ⁠, so

\begin{aligned} &\frac{0.9979980 - 1.0019980}{0.002} \\ &\approx -2.000, \end{aligned}

matching $y' = -2$ ⁠.

Common mistakes

Forgetting $y'$ on the right: $\frac{d}{dx}\tan^{-1}y = \frac{y'}{1 + y^2}$ ⁠, not $\frac{1}{1 + y^2}$ ⁠.
Differentiating $x$ to 0 or leaving it as $x$ ⁠: it gives 1.
Expanding $(1 + y^2)(1 + y')$ with a term missing. There are four terms.

What to take away

Implicit check: differentiate, clear fractions, simplify to the given equation.
Find where $y'$ fails to exist; that is where the verification stops.
Compare Q7 and Q8.

Every fact used, and where it comes from

Fact	Used in	Source
$(\tan^{-1}y)' = \frac{1}{1 + y^2}$ ⁠; chain rule	Step 1	Chapter 5, Sections 5.3.1 and 5.3.3 (Basics)
Implicit differentiation	Step 1	Chapter 5, Section 5.3.2
A solution makes L.H.S. = R.H.S.	Steps 3, 4	Section 9.3 (Lesson)
Order and degree (polynomial condition); particular solution	Step 4	Sections 9.2.1, 9.2.2 and 9.3 (Lesson)

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Exercise 9.2 · Q9: x+y=tan⁡−1yx + y = \tan^{-1}yx+y=tan−1y solves y2y′+y2+1=0y^2y' + y^2 + 1 = 0y2y′+y2+1=0

Basics you need

Solution

Differentiate both sides with respect to xxx

Clear the fraction

Expand and simplify

Where y′y'y′ exists, and conclusion

Every fact used, and where it comes from

Exercise 9.2 · Q9: $x + y = \tan^{-1}y$ solves $y^2y' + y^2 + 1 = 0$

Differentiate both sides with respect to $x$

Where $y'$ exists, and conclusion