Differential-Equations · Unit 1 · Video 3 · Interactive Practice

Worked Example: $y' = -x/y$ , Perpendicular Isoclines, and Circles

IKey Formulas

Formula	Name	What you need
$-\dfrac{x}{y} = C \;\Longrightarrow\; y = -\dfrac{1}{C}\,x$	Isocline of slope $C$ — a ray through the origin	A slope value $C \neq 0$ ; the unsolved form $Cy = -x$ reaches $C = 0$
$\left(-\dfrac{1}{C}\right)\cdot C = -1$	Isocline slope times element slope	Negative reciprocals, so the two are perpendicular
$y\,dy = -x\,dx \;\Longrightarrow\; x^2 + y^2 = c_1^2$	Integral curves, by separation	Integrate both sides; $2c = c_1^2$ because the left side is $\ge 0$
$y = \sqrt{c_1^2 - x^2},\quad -c_1 < x < c_1$	Explicit solution and its domain	One point on the curve, to fix $c_1$

Key Insight: Because every isocline carries elements perpendicular to itself, an integral curve must cross every ray through the origin at a right angle — and only circles centred at the origin do that. The circles arrive before any integration; the domain $-c_1 < x < c_1$ arrives only after it.

IIVisualization 1 — Why the Elements Stand Perpendicular

The isocline of slope $C$ is the ray $y = -\tfrac{1}{C}x$ , and the two slopes multiply to $-1$ .

Isocline direction 135°

💡 The solved form $y = -\tfrac{1}{C}x$ can never produce $C = 0$ — it divides by it. Written unsolved as $Cy = -x$ , the value $C = 0$ gives $x = 0$ : the $y$ -axis is a genuine isocline, carrying horizontal elements.

IIIVisualization 2 — Following the Elements

A curve that follows the elements crosses every ray through the origin at a right angle.

t 0.00 Start

IVVisualization 3 — Separating Variables, and the Domain

Separation confirms the circles, then hands you a domain the equation itself never mentions.

Step 1 c₁ 3.0

1 · separate the variables

\frac{dy}{dx} = -\frac{x}{y}

y\,dy = -x\,dx

2 · integrate both sides

\int y\,dy = -\int x\,dx

\frac{y^2}{2} = -\frac{x^2}{2} + c

3 · name the constant

x^2 + y^2 = 2c \equiv c_1^2

The left side is $\ge 0$ , so the constant is a square.

4 · solve for y — two branches

y = \sqrt{c_1^2 - x^2}

y = -\sqrt{c_1^2 - x^2}

5 · the domain

c_1^2 - x^2 > 0

-c_1 < x < c_1

6 · check, implicitly

2x + 2y\,y' = 0

y' = -\frac{x}{y}\ \checkmark

💡 Here the domain can be read off the explicit solution. For most equations it cannot: how far a solution extends is discovered by computing it, never by inspecting the equation.

VQuiz Questions

Problem 1 · Finding an Isocline

Given: $y' = -\dfrac{x}{y}$ — find the isocline along which every line element has slope $C = 2$ .

✅ Correct! Setting

-x/y = 2

gives

y = -\tfrac{1}{2}x

. Its own slope is

-\tfrac{1}{2}

, and

\left(-\tfrac{1}{2}\right)(2) = -1

— the elements stand perpendicular to the isocline that carries them.

❌ That is the elements' slope drawn as the isocline. On

y = 2x

the equation gives

-x/(2x) = -\tfrac{1}{2}

, so

y = 2x

is the isocline of

C = -\tfrac{1}{2}

, not of

C = 2

❌ Sign slip. On

y = \tfrac{1}{2}x

the elements have slope

-x/\left(\tfrac{1}{2}x\right) = -2

. Solving

-x/y = 2

puts the minus sign on

y

-x = 2y

, so

y = -\tfrac{1}{2}x

❌ You used $y = -Cx$ instead of $y = -\tfrac{1}{C}x$ . On

y = -2x

the equation gives

-x/(-2x) = \tfrac{1}{2}

, so that line is the isocline of

C = \tfrac{1}{2}

❌ Not quite. Set the right-hand side equal to the slope you want and solve for

y

-x/y = 2

Show solution

An isocline is the set of points where the equation prescribes one fixed slope, so set the right-hand side equal to that slope and solve for $y$ :

-\frac{x}{y} = 2 \quad\Longrightarrow\quad -x = 2y \quad\Longrightarrow\quad y = -\frac{1}{2}x

Check a point. $(2, -1)$ lies on this line, and there $y' = -2/(-1) = 2\ \checkmark$ .

Check the geometry. The isocline's own slope is $-\tfrac{1}{2}$ and the elements it carries have slope $2$ ; the product is $-1$ , which is the general fact $\left(-\tfrac{1}{C}\right)\cdot C = -1$ for this equation.

Problem 2 · The Isocline's Slope Is Not the Slope It Carries

Given: the line $y = 3x$ is an isocline of $y' = -\dfrac{x}{y}$ — find the constant slope $C$ carried by the line elements along it.

✅ Correct! Every point of the line is

(a, 3a)

, and there

y' = -a/(3a) = -\tfrac{1}{3}

. The two slopes check out:

3 \cdot \left(-\tfrac{1}{3}\right) = -1

❌ That is the isocline's own slope. An isocline is only a guide curve; the slope it carries is generally different from its own. For this equation the two are as different as possible — their product is forced to be

-1

❌ Right size, wrong sign.

\tfrac{1}{3}

is the plain reciprocal, and

3 \cdot \tfrac{1}{3} = 1

. Perpendicularity needs the negative reciprocal,

-\tfrac{1}{3}

❌ Negative, but not reciprocal.

3 \cdot (-3) = -9

, not

-1

. Evaluating the equation directly at

(1, 3)

gives

-1/3

❌ Not quite. Evaluate

-x/y

at any point of the line —

(1, 3)

will do.

Show solution

Directly. Every point of $y = 3x$ has the form $(a, 3a)$ with $a \neq 0$ , so the equation gives

y' = -\frac{x}{y} = -\frac{a}{3a} = -\frac{1}{3}

The same value at every point — which is what makes the line an isocline.

From the general form. The isocline of slope $C$ is $y = -\tfrac{1}{C}x$ , so its own slope is $-1/C$ . Here that must equal $3$ :

-\frac{1}{C} = 3 \quad\Longrightarrow\quad C = -\frac{1}{3}

And $3 \cdot \left(-\tfrac{1}{3}\right) = -1$ , so the elements meet this isocline at a right angle, exactly as they do on every other one.

Problem 3 · The Solution Through a Point, and How Far It Goes

Given: the solution of $y' = -\dfrac{x}{y}$ passing through $(3, 4)$ — find it explicitly, and find its domain.

Which explicit solution?

What is its domain?

✅ Correct!

c_1^2 = 3^2 + 4^2 = 25

; the point has

y = 4 > 0

, so the upper branch; and

25 - x^2 > 0

confines the solution to

-5 < x < 5

. Nothing in

y' = -x/y

announced that interval.

❌ Wrong branch. At

x = 3

this gives

y = -4

, not

4

. The negative square root is the lower semicircle;

(3, 4)

sits on the upper one.

❌ You subtracted where the circle adds. The relation is

x^2 + y^2 = c_1^2

, so

c_1^2 = 9 + 16 = 25

, not

16 - 9 = 7

. With

7

under the root the expression is not even real at

x = 3

❌ Sign inside the root. From

x^2 + y^2 = c_1^2

comes

y^2 = c_1^2 - x^2

, so

x^2

is subtracted. As a check,

\sqrt{25 + 9} = \sqrt{34} \approx 5.83 \neq 4

❌ Check the explicit form. Separating gives

x^2 + y^2 = c_1^2

; substitute

(3, 4)

to fix

c_1

, then solve for

y

on the branch that contains the point.

❌ This is exactly the assumption to avoid.

\sqrt{25 - x^2}

is undefined for

|x| > 5

, and at

x = \pm 5

the circle turns vertical with

y = 0

, where

-x/y

has no value at all.

❌ Only half of the interval. Nothing restricts the solution to positive

x

: the upper semicircle runs continuously from

x = -5

x = 5

, passing through

(0, 5)

❌ Those bounds come from the $y$ -coordinate. The interval is set by the radius

c_1 = 5

, not by

y = 4

: the condition is

c_1^2 - x^2 > 0

❌ Check the domain. The square root demands

25 - x^2 > 0

; solve that inequality for

x

Show solution

Step 1 — separate and integrate.

y\,dy = -x\,dx \quad\Longrightarrow\quad \frac{y^2}{2} = -\frac{x^2}{2} + c \quad\Longrightarrow\quad x^2 + y^2 = c_1^2

Step 2 — fix the constant with the point. Substituting $(3, 4)$ :

c_1^2 = 3^2 + 4^2 = 25, \qquad c_1 = 5

Step 3 — pick the branch. The point has $y = 4 > 0$ , so the solution is the upper semicircle:

y = \sqrt{25 - x^2}

Step 4 — the domain. The square root requires $25 - x^2 > 0$ , that is $x^2 < 25$ :

-5 < x < 5

The interval is open at both ends. At $x = \pm 5$ the solution would have $y = 0$ , where $-x/y$ is undefined and the circle is vertical, so the solution cannot be extended past those points. Verify the equation itself:

y' = \frac{-x}{\sqrt{25 - x^2}} = -\frac{x}{y}\ \checkmark

The equation is identical for every solution of this family; only the constant $c_1$ — supplied by the point, not by the equation — decides how wide the interval is.

Problem 4 · When the Isoclines Are Not Perpendicular

Given: a different equation, $y' = \dfrac{y}{x}$ , whose isocline of slope $C$ is the line $y = Cx$ — find the product of the isocline's slope with the slope it carries, and find the integral curves.

Isocline slope × element slope

So the integral curves are

✅ Correct! The line

y = Cx

has slope

C

and carries elements of slope

y/x = C

: the product is

C^2

, never

-1

. The elements lie along their isocline, so each isocline is its own integral curve.

❌ That is the $y' = -x/y$ answer, imported. Perpendicular isoclines are a property of that particular right-hand side. Here both slopes equal

C

, so the product is

C^2 \ge 0

and can never be

-1

for a real

C

❌ True only when $C = \pm 1$ . In general the product is

C \cdot C = C^2

; for

C = 2

it is

4

❌ No minus sign enters. Both slopes are

+C

: the isocline's own slope is

C

, and the elements it carries have slope

y/x = C

as well, so the product is

C^2

❌ Check both slopes. Read one off the line

y = Cx

, and get the other by evaluating

y/x

on that same line.

❌ Those solve the other equation. Circles centred at the origin satisfy

y' = -x/y

. For

y' = y/x

the slope at

(x, y)

points away from the origin, not around it.

❌ Those belong to $y' = x/y$ . Differentiating

y^2 - x^2 = k

implicitly gives

2y\,y' - 2x = 0

, so

y' = x/y

— a third equation, with its own geometry.

❌ A vertical line has no slope anywhere. On

x = k

(with

k \neq 0

) the equation prescribes the finite slope

y/k

at every point, so that line cannot be an integral curve.

❌ Reconsider. If the elements on an isocline are parallel to it, the isocline follows its own elements everywhere — which is the definition of an integral curve.

Show solution

The two slopes. The line $y = Cx$ has slope $C$ . At any point $(a, Ca)$ on it the equation gives

y' = \frac{y}{x} = \frac{Ca}{a} = C

so the elements have slope $C$ as well and the product is $C \cdot C = C^2$ . Since $C^2 \ge 0$ , it is never $-1$ : these isoclines are never perpendicular to their elements. They are parallel to them.

What that forces. An integral curve follows the elements. If the elements along a line point straight down that line, the line already follows them, so each isocline is an integral curve. Separating confirms it:

\frac{dy}{y} = \frac{dx}{x} \quad\Longrightarrow\quad \ln|y| = \ln|x| + k \quad\Longrightarrow\quad y = kx

The contrast. For $y' = -x/y$ the slopes multiply to $-1$ : elements perpendicular to the rays, integral curves circular. For $y' = y/x$ they multiply to $C^2$ : elements along the rays, integral curves the rays. In both cases the isocline geometry alone settles the answer — and in both cases the $y$ -axis or $x$ -axis has to be excluded where the right-hand side blows up.

Solved: 0 / 4

Worked Example: y′=−x/yy' = -x/yy′=−x/y, Perpendicular Isoclines, and Circles

IKey Formulas

IIVisualization 1 — Why the Elements Stand Perpendicular

IIIVisualization 2 — Following the Elements

IVVisualization 3 — Separating Variables, and the Domain

VQuiz Questions

Worked Example: $y' = -x/y$ , Perpendicular Isoclines, and Circles