Differential-Equations · Unit 1 · Video 6 · Interactive Practice

When Existence and Uniqueness Fail: A Fan of Lines Through One Point

IKey Formulas

Formula	Name	What it tells you
$x\,\dfrac{dy}{dx} = y - 1$	The equation	Not yet in standard form
$y = 1 + Cx$	General solution	A line of slope $C$ through $(0,1)$
$y' = f(x,y) = \dfrac{y-1}{x}$	Standard form	$f$ is undefined on the line $x = 0$
$f_y = \dfrac{1}{x}$	Uniqueness hypothesis	Undefined on the same line

Key Insight: The theorem is not violated — it was never speaking. Its hypotheses ask about $f$ and $f_y$ in standard form, and on $x = 0$ neither exists. With no promise in force the count is free: no solution through $(0,b)$ for $b \neq 1$ , and every solution through $(0,1)$ .

IISeparating the Variables

Separation costs two lines of the plane; the general solution hands one of them back.

Step 1

Step 1 — Separate the variables

\frac{dy}{y-1} = \frac{dx}{x}

Dividing by

y-1

and by

x

assumes

y \neq 1

and

x \neq 0

: two lines are set aside before any integration happens.

Step 2 — Integrate both sides

\ln\left|y-1\right| = \ln\left|x\right| + c_1

The absolute values are not decoration —

\ln

accepts only positive arguments, and neither

y-1

nor

x

has a fixed sign.

Step 3 — Exponentiate, then absorb the constant

\left|y-1\right| = e^{c_1}\left|x\right|

For

e^{c_1} = 1

this is a pair of lines,

y = 1+x

and

y = 1-x

. Setting

C = \pm e^{c_1}

lets one formula carry both branches:

y - 1 = Cx

Step 4 — The general solution

y = 1 + Cx

The slope is

C

and can be anything; the value at

x = 0

1

no matter what

C

is. One hub, every slope.

Step 5 — Check, and recover the lost line

y' = C \quad\Longrightarrow\quad x\,y' = Cx = y - 1 \ \checkmark

C = 0

gives

y = 1

, which makes both sides zero: the line divided away in Step 1 returns as an ordinary member of the family.

IIICounting Solutions Through a Point

Through each point of the plane, how many members of $y = 1 + Cx$ pass?

x₀ 2 y₀ 3

💡 Both failures sit on the same line $x = 0$ — and no theorem is broken there, for the reason the next section makes visible.

IVStandard Form and the Bad Points

The theorem speaks only in standard form, and only where $f(x,y)$ actually exists.

Equation

💡 The habit this buys: write $y' = f(x,y)$ first, then mark every point where $f$ or $f_y$ fails to be continuous — those are the only places where a guarantee can go missing.

VQuiz Questions

Problem 1 · Pick the Member

Given: the general solution $y = 1 + Cx$ of $x\,y' = y - 1$ — find the member that passes through $(-2, 4)$ .

✅ Correct!

C = \dfrac{4-1}{-2} = -\dfrac{3}{2}

, and

1 - \tfrac{3}{2}(-2) = 1 + 3 = 4

❌ Not quite. That is

C = \dfrac{y_0}{x_0}

. The hub sits at height

1

, not

0

, so the rise from the hub is

y_0 - 1

, not

y_0

❌ Close, but check the sign. Here

x_0 = -2

is negative while

y_0 - 1 = 3

is positive, so

C

must be negative.

❌ Not quite. Substitute the point into

y_0 = 1 + Cx_0

and solve for

C

4 = 1 + C(-2)

Show solution

Every member has $y = 1$ at $x = 0$ , so the point fixes the slope. Substitute $(x_0, y_0) = (-2, 4)$ :

y_0 = 1 + Cx_0 \quad\Longrightarrow\quad 4 = 1 + C(-2)

C = \frac{y_0 - 1}{x_0} = \frac{4 - 1}{-2} = -\frac{3}{2}

So the member is $y = 1 - \tfrac{3}{2}x$ .

Verify: at $x = -2$ , $\ y = 1 - \tfrac{3}{2}(-2) = 1 + 3 = 4\ \checkmark$ ; and $x\,y' = (-2)\left(-\tfrac{3}{2}\right) = 3 = y - 1\ \checkmark$ .

Problem 2 · Standing on the Axis

Given: $x\,y' = y - 1$ — how many solutions pass through the point $(0, -1)$ ?

✅ Correct! Setting

x = 0

in the equation gives

0 = y - 1

, so any solution defined at

x = 0

must have

y = 1

there. Height

-1

is unreachable.

❌ That is the hub, not this point. Every member does meet at

(0, 1)

, but at

x = 0

they all take the value

1 + C(0) = 1

— none of them reaches

-1

❌ Check that it solves the equation. For

y = -1

x\,y' = 0

while

y - 1 = -2

, so the two sides disagree. The only constant solution is

y = 1

❌ Not quite. Every member of

y = 1 + Cx

takes the value

1

x = 0

, whatever

C

is.

Show solution

From the family: at $x = 0$ , $\ y = 1 + C(0) = 1$ for every $C$ . No member reaches height $-1$ .

From the equation itself (this rules out solutions outside the family too): put $x = 0$ into $x\,y' = y - 1$ :

0 \cdot y'(0) = y(0) - 1 \quad\Longrightarrow\quad 0 = y(0) - 1 \quad\Longrightarrow\quad y(0) = 1

A solution through $(0,-1)$ would need $y(0) = -1$ and $y(0) = 1$ at once. There is no solution: existence fails at $(0,-1)$ , and by the same argument at every $(0,b)$ with $b \neq 1$ .

Problem 3 · Consulting the Theorem Correctly

Given: $x\,y' = y - 1$ — rewrite it as $y' = f(x,y)$ and locate the points where the existence-and-uniqueness hypotheses fail.

What is $f(x,y)$ ?

Where do the hypotheses fail?

✅ Correct!

f = \dfrac{y-1}{x}

and

f_y = \dfrac{1}{x}

are both undefined exactly when

x = 0

, so the theorem says nothing along the entire

y

-axis — including the hub.

❌ Check the algebra. Standard form isolates

y'

alone on the left, so divide

x\,y' = y - 1

through by

x

— do not multiply.

❌ Look at the denominator. The hypotheses ask that

f

and

f_y

be continuous near the point; both are quotients with

x

underneath, so both die on the whole line

x = 0

— not at one isolated point.

Show solution

Step 1: Standard form. Divide by $x$ to leave the derivative alone on the left:

y' = f(x,y) = \frac{y-1}{x}

Step 2: Test the hypotheses. Existence needs $f$ continuous near the point; uniqueness needs $f_y$ continuous as well:

f_y = \frac{\partial}{\partial y}\left(\frac{y-1}{x}\right) = \frac{1}{x}

For $x \neq 0$ both are continuous, and the picture obeys: exactly one line through each off-axis point.

Step 3: Read off the failure set. At $x = 0$ neither $f$ nor $f_y$ is even defined, so continuity never comes up for discussion. The hypotheses fail at every point of the line $x = 0$ , and with no promise in force the count comes out however it pleases: $0$ solutions through $(0,b)$ for $b \neq 1$ , infinitely many through $(0,1)$ .

Note that $\dfrac{1}{x}$ is $f_y$ , not $f$ — a tempting swap once both expressions are on the page.

Problem 4 · A New Equation, Same Habit

Given: $(y-2)\,y' = x - 1$ — find the set of points at which the existence-and-uniqueness theorem makes no promise.

✅ Correct! Standard form is

y' = \dfrac{x-1}{y-2}

, and the denominator vanishes precisely on

y = 2

❌ Wrong line.

x = 1

makes the numerator zero, which simply gives the harmless slope

y' = 0

. It is the vanishing denominator that destroys

f

❌ Too small a set. At

(1,2)

the quotient is

0/0

, but at

(5,2)

it is

4/0

— undefined as well. Every point with

y = 2

is a bad point.

❌ Not quite. A quotient of polynomials is continuous only where its denominator is nonzero. Solve

y - 2 = 0

Show solution

Step 1: Write $y' = f(x,y)$ . Divide by $y - 2$ :

y' = f(x,y) = \frac{x-1}{y-2}

Step 2: Mark where $f$ fails to exist. The denominator vanishes when $y = 2$ , so $f$ is undefined at every point of the horizontal line $y = 2$ — including $(1,2)$ , where the quotient reads $0/0$ .

Step 3: Check the uniqueness hypothesis too.

f_y = -\frac{x-1}{(y-2)^2}

which dies on the same line. Off that line both $f$ and $f_y$ are continuous, so the theorem delivers exactly one solution curve through each such point.

This is the same shape of failure as $x\,y' = y - 1$ (bad set $x = 0$ ) and as the circle equation $y' = -\dfrac{x}{y}$ (bad set $y = 0$ ): a denominator that vanishes along a whole line.

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