Chapter 9 · Differential Equations

Exercise 9.2 · Q5: $y = Ax$ solves $xy' = y$

Exercise 9.2Verifying a 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:

5. $y = Ax$ : $xy' = y$ $(x \ne 0)$

Exam answer

Given $y = Ax$ ⁠. Differentiating with respect to $x$ ⁠,

y' = A

Substituting $y'$ in the L.H.S. of the given equation,

\begin{aligned} \text{L.H.S.} &= xy' = x \cdot A \\ &= Ax = y = \text{R.H.S.} \end{aligned}

$\therefore$ $y = Ax$ is a solution of the given differential equation $(x \ne 0)$ ⁠. Hence verified.

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

The one idea this problem is built on

The graph of $y = Ax$ is a straight line through the origin with slope $A$ ⁠. Multiplying the slope by $x$ gives back $Ax$ ⁠, the height of the line. So $xy' = y$ for every line through the origin, whatever its slope $A$ ⁠.

Given, to find, and the plan

Given: The function $y = Ax$ ⁠, where $A$ is an arbitrary constant; the equation $xy' = y$ ⁠, with the condition $x \ne 0$ ⁠.
To show: That $y = Ax$ satisfies the equation for every allowed $x$ and every $A$ ⁠.
Plan: (1) Find $y'$ ⁠. (2) Substitute into the left side and compare with the right side. (3) Discuss the condition $x \ne 0$ ⁠. (4) Count the constants.

Basics you need

1What a solution is

A function is a solution when substituting it for $y$ makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

2Derivative of $Ax$

For a constant $A$ ⁠, $\frac{d}{dx}(Ax) = A$ ⁠: the slope of a straight line (Chapter 5). Basics: y′ and y″

3Arbitrary constants

A general solution carries as many arbitrary constants as the order (Summary, Chapter 9). Basics: order n gives n constants

Solution

Differentiate
$y = Ax \;\Rightarrow\; y' = A.$
$A$ is a constant factor, so it stays in front, and $\frac{d}{dx}x = 1$ ⁠.

Constant multiple rule (Chapter 5). Basics: y′ and y″
Substitute and compare
$\begin{aligned} \text{L.H.S.} &= xy' = x \cdot A = Ax, \\ \text{R.H.S.} &= y = Ax. \end{aligned}$
So L.H.S. = R.H.S.

Replace $y'$ by $A$ on the left and $y$ by $Ax$ on the right. Both sides become $Ax$ ⁠.

Section 9.3: substitute and compare the two sides. Lesson: is it a solution? Examples 2 and 3
The condition $x \ne 0$

No step divides, so the equality $xy' = y$ holds for every real $x$ ⁠; at $x = 0$ both sides are simply 0. The book adds $x \ne 0$ because that is where the equation can be divided by $x$ and read as a slope rule, $y' = \frac{y}{x}$ ⁠. On $x \ne 0$ the check holds as shown, so the function is a solution there: hence verified.

The condition belongs to the equation, not to our working. Saying where each step is valid is part of a complete verification.

A solution satisfies the equation on its domain (Section 9.3). Lesson: is it a solution?
One constant: a general solution

The equation has order 1 and degree 1 (it is a polynomial equation in $y'$ ⁠, with $y'$ to the power 1). The function $y = Ax$ has one arbitrary constant, as many as the order, so it is a general solution. Each value of $A$ is a particular solution: $A = 2$ gives $y = 2x$ ⁠, $A = 0$ gives $y = 0$ ⁠.

Geometrically, $xy' = y$ says the tangent at each point passes through the origin. On an interval where $x \ne 0$ ⁠, every solution is a piece of such a line, because $\left(\frac{y}{x}\right)' = \frac{xy' - y}{x^2} = 0$ there, so $\frac{y}{x}$ is a constant.

General and particular solutions (Section 9.3 and the Summary). Basics: order n gives n constants

Result

With $y' = A$ ⁠, L.H.S. $= xy' = Ax = y =$ R.H.S. for every $x \ne 0$ (in fact for every real $x$ ⁠) and every constant $A$ ⁠. Hence $y = Ax$ is a solution of $xy' = y$ ⁠, its general solution. (The answer key gives no answer for this question, since it is a verification.)

Check it with numbers

Take $A = 3$ and $x = 2$ ⁠: $y = 6$ ⁠, $y' = 3$ ⁠, and $xy' = 2 \times 3 = 6 = y$ ⁠.

Take $A = -0.5$ and $x = -4$ ⁠: $y = 2$ ⁠, $y' = -0.5$ ⁠, and $xy' = (-4)(-0.5) = 2 = y$ ⁠.

Common mistakes

Substituting $y = Ax$ on both sides but forgetting to replace $y'$ by $A$ ⁠.
Treating $A$ as a variable and differentiating it. It is an arbitrary constant.
Ignoring the condition: state where the check holds. Here it is all $x \ne 0$ (and $x = 0$ as well).

What to take away

$xy' = y$ is the equation of all lines through the origin.
Keep $A$ as a letter so the check covers the whole family; compare Q2.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}(Ax) = A$ for a constant $A$	Step 1	Chapter 5 (Basics)
A solution makes L.H.S. = R.H.S. when substituted	Steps 2, 3	Section 9.3 (Lesson)
Order and degree (polynomial condition)	Step 4	Sections 9.2.1 and 9.2.2 (Lesson)
General solution has as many constants as the order	Step 4	Section 9.3 and the Summary of Chapter 9 (Basics)

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Exercise 9.2 · Q5: y=Axy = Axy=Ax solves xy′=yxy' = yxy′=y

Basics you need

Solution

Differentiate

Substitute and compare

The condition x≠0x \ne 0x=0

One constant: a general solution

Every fact used, and where it comes from

Exercise 9.2 · Q5: $y = Ax$ solves $xy' = y$

The condition $x \ne 0$