Chapter 9 · Differential Equations

Exercise 9.2 · Q10: $y = \sqrt{a^2 - x^2}$ solves $x + y\frac{dy}{dx} = 0$

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:

10. $y = \sqrt{a^2 - x^2}$ ⁠, $x \in (-a, a)$ : $x + y\dfrac{dy}{dx} = 0$ $(y \ne 0)$

Exam answer

Given $y = \sqrt{a^2 - x^2}$ ⁠, $x \in (-a, a)$ ⁠. Differentiating with respect to $x$ ⁠,

\begin{aligned} \frac{dy}{dx} &= \frac{-2x}{2\sqrt{a^2 - x^2}} \\ &= \frac{-x}{\sqrt{a^2 - x^2}} \end{aligned}

Substituting $y$ and $\dfrac{dy}{dx}$ in the L.H.S.,

\begin{aligned} \text{L.H.S.} &= x + y\frac{dy}{dx} \\ &= x + \sqrt{a^2 - x^2} \\ &\quad \times \frac{-x}{\sqrt{a^2 - x^2}} \\ &= x - x = 0 = \text{R.H.S.} \end{aligned}

$\therefore$ $y = \sqrt{a^2 - x^2}$ is a solution of the given differential equation $(x \in (-a, a))$ ⁠. Hence verified.

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

The one idea this problem is built on

The graph of $y = \sqrt{a^2 - x^2}$ is the upper half of the circle $x^2 + y^2 = a^2$ ⁠. Differentiating gives $y\frac{dy}{dx} = -x$ ⁠, which is exactly the equation $x + y\frac{dy}{dx} = 0$ ⁠. In pictures: the radius to a point $(x, y)$ has slope $\frac{y}{x}$ ⁠, the tangent has slope $-\frac{x}{y}$ ⁠, and the two are perpendicular.

Given, to find, and the plan

Given: The function $y = \sqrt{a^2 - x^2}$ for $x \in (-a, a)$ (here $a > 0$ ⁠, so that the interval is not empty); the equation $x + y\frac{dy}{dx} = 0$ ⁠, with $y \ne 0$ ⁠.
To show: That the function satisfies the equation at every $x$ in $(-a, a)$ ⁠.
Plan: (1) Differentiate by the chain rule. (2) Substitute into the left side and simplify to 0. (3) Check that every step is valid on $(-a, a)$ ⁠. (4) Interpret, and conclude.

Basics you need

1Chain rule for a square root

$\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$ for $u > 0$ (Section 5.3.1). Basics: y′ and y″

2What a solution is

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

3Slope of a tangent

$\frac{dy}{dx}$ at a point is the slope of the tangent there; perpendicular lines have slopes whose product is $-1$ ⁠. Basics: a little arrow at every point

Solution

Differentiate by the chain rule
$\begin{aligned} y &= (a^2 - x^2)^{1/2} \\ \frac{dy}{dx} &= \tfrac{1}{2}(a^2 - x^2)^{-1/2}\cdot(-2x) \\ &= \frac{-x}{\sqrt{a^2 - x^2}}. \end{aligned}$
Outer function $\sqrt{u}$ ⁠, derivative $\frac{1}{2\sqrt{u}}$ ⁠; inner function $u = a^2 - x^2$ ⁠, derivative $-2x$ ( $a$ is a constant). Multiply, and the 2s cancel.

Chain rule (Section 5.3.1). Basics: y′ and y″
Substitute into the left side
$\begin{aligned} \text{L.H.S.} &= x + y\frac{dy}{dx} \\ &= x + \sqrt{a^2 - x^2}\cdot\frac{-x}{\sqrt{a^2 - x^2}} \\ &= x - x = 0 = \text{R.H.S.} \end{aligned}$
The factor $\sqrt{a^2 - x^2}$ from $y$ cancels the same factor in the denominator of $\frac{dy}{dx}$ ⁠, leaving $-x$ ⁠, which cancels $x$ ⁠.

Section 9.3: substitute and compare the two sides. Lesson: is it a solution? Examples 2 and 3
Every step is valid on $(-a, a)$

For $-a < x < a$ we have $x^2 < a^2$ ⁠, so $a^2 - x^2 > 0$ ⁠. Hence the root is defined and differentiable, $y = \sqrt{a^2 - x^2} > 0$ (so $y \ne 0$ ⁠, the printed condition), and the denominator in Step 1 is not 0, so the cancellation in Step 2 is allowed. At $x = \pm a$ ⁠, $y = 0$ and $\frac{dy}{dx}$ does not exist (vertical tangents), which is why the interval is open.

The root and the division are the two places where a step could fail; the open interval $(-a, a)$ is exactly where both are safe.

A solution satisfies the equation on its domain (Section 9.3). Lesson: Exercise 9.2 player
Meaning, and conclusion

Squaring, $y^2 = a^2 - x^2$ ⁠, so the graph is the upper half of the circle $x^2 + y^2 = a^2$ ⁠. The equation says $\frac{dy}{dx} = -\frac{x}{y}$ ⁠: the tangent slope is $-\frac{x}{y}$ ⁠, while the radius from the origin has slope $\frac{y}{x}$ ⁠; their product is $-1$ (for $x \ne 0$ ⁠), so the tangent is perpendicular to the radius, as it must be for a circle.

Hence verified: $y = \sqrt{a^2 - x^2}$ is a solution on $(-a, a)$ ⁠. The equation has order 1 and degree 1 (a polynomial equation in $\frac{dy}{dx}$ ⁠, power 1). Treating $a$ as an arbitrary constant, the function has one constant, as many as the order: it gives the upper halves of all circles centred at the origin.

Seeing the geometry explains why the check had to work: every circle centred at the origin has its tangents perpendicular to its radii.

$\frac{dy}{dx}$ is the slope of the tangent. Basics: a little arrow at every point

The upper half of

x^2 + y^2 = 25

(

a = 5

⁠), drawn with equal scales. At

P(3, 4)

the radius has slope

\frac{4}{3}

and the tangent has slope

\frac{dy}{dx} = -\frac{3}{4}

⁠; they meet at a right angle. The open circles at

x = \pm 5

mark the ends excluded from

(-a, a)

⁠.

Result

With $\frac{dy}{dx} = \frac{-x}{\sqrt{a^2 - x^2}}$ ⁠, L.H.S. $= x + y\frac{dy}{dx} = x - x = 0 =$ R.H.S. for every $x \in (-a, a)$ ⁠, where $y > 0$ ⁠. Hence $y = \sqrt{a^2 - x^2}$ is a solution of $x + y\frac{dy}{dx} = 0$ ⁠. (The answer key gives no answer for this question, since it is a verification.)

Check it with numbers

Take $a = 5$ and $x = 3$ ⁠: $y = \sqrt{25 - 9} = 4$ and $\frac{dy}{dx} = \frac{-3}{4} = -0.75$ ⁠. Then $x + y\frac{dy}{dx} = 3 + 4(-0.75) = 0$ ⁠.

Take $a = 5$ and $x = -4$ ⁠: $y = \sqrt{25 - 16} = 3$ and $\frac{dy}{dx} = \frac{4}{3}$ ⁠. Then $x + y\frac{dy}{dx} = -4 + 3 \times \frac{4}{3} = 0$ ⁠.

Common mistakes

Forgetting the inner derivative $-2x$ , or losing its minus sign.
Treating $a$ as a variable. It is a constant, so $\frac{d}{dx}a^2 = 0$ ⁠.
Ignoring the endpoints. At $x = \pm a$ ⁠, $y = 0$ and $\frac{dy}{dx}$ does not exist.

What to take away

$x + y\frac{dy}{dx} = 0$ is the equation of circles centred at the origin: tangent perpendicular to radius.
State where the root is positive; that is the domain of the verification. Compare Q4 (root always positive) and Q6 (root of a square).

Every fact used, and where it comes from

Fact	Used in	Source
Chain rule; $\frac{d}{dx}\sqrt{u} = \frac{u'}{2\sqrt{u}}$	Step 1	Chapter 5, Section 5.3.1 (Basics)
A solution makes L.H.S. = R.H.S. when substituted	Steps 2, 3	Section 9.3 (Lesson)
Perpendicular slopes multiply to $-1$	Step 4	Class XI coordinate geometry
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 · Q10: y=a2−x2y = \sqrt{a^2 - x^2}y=a2−x2​ solves x+ydydx=0x + y\frac{dy}{dx} = 0x+ydxdy​=0

Basics you need

Solution

Differentiate by the chain rule

Substitute into the left side

Every step is valid on (−a,a)(-a, a)(−a,a)

Meaning, and conclusion

Every fact used, and where it comes from

Exercise 9.2 · Q10: $y = \sqrt{a^2 - x^2}$ solves $x + y\frac{dy}{dx} = 0$

Every step is valid on $(-a, a)$