Differential-Equations · Unit 2 · Video 7 · Interactive Practice

The Wall the Equation Never Mentions: $y' = y^2$ and Its Private Singularities

IKey Formulas

Formula	Name	What it tells you
$y' = y^2$	The equation	Separable; no $x$ on the right
$y = \dfrac{1}{c - x}$	General solution	One curve for each constant $c$
$y(0) = y_0 \;\Longrightarrow\; c = \dfrac{1}{y_0}$	Constant from the initial condition	Which curve you are on
$x = c$	Singular point (the wall)	Where the denominator vanishes and the solution ends

Key Insight: The equation contains no $c$ . The wall's position enters only through the initial condition, so a method that reads nothing but slopes cannot see it coming. The single exception is $y \equiv 0$ , the solution lost when we divided by $y^2$ — the only one with no wall.

IISeparating the Variables

Separation costs one solution and produces a constant that turns out to locate a wall.

Step 1

Step 1 — Write the derivative as a quotient

\frac{dy}{dx} = y^2

The right-hand side carries no $x$ : the slope at a point depends on its height alone, so the field is the same in every column.

Step 2 — Separate the variables

\frac{dy}{y^2} = dx

Dividing by $y^2$ assumes $y \neq 0$ , and $y \equiv 0$ does solve the equation. One solution slips away here.

Step 3 — Integrate both sides

-\frac{1}{y} = x - c

$\int y^{-2}\,dy = -\dfrac{1}{y}$ and $\int dx = x + \text{constant}$ . The arbitrary constant is written as $-c$ , which saves a sign in the next line.

Step 4 — Solve for y

\frac{1}{y} = c - x \qquad\Longrightarrow\qquad y = \frac{1}{c - x}

One curve for each value of $c$ — the whole family, and not complicated at all.

Step 5 — Locate the wall

c - x = 0 \qquad\Longrightarrow\qquad x = c

Each member blows up at its own $x = c$ and stops existing there. Only $y \equiv 0$ , which is not a member of the family, runs on forever.

IIIWhere the Wall Stands

The starting height alone fixes the constant, $c = 1/y(0)$ , and the solution ends at $x = c$ .

y(0) 1

IVMarching Into the Wall

RK4 tracks the true solution to four decimals, then keeps marching where no solution exists.

Step n 0

$x$	RK4, $h = 0.01$ , from $(0,1)$	$\dfrac{1}{1-x}$
$0$	—	$1$
$0.5$	—	$2$
$0.9$	—	$10$
$0.99$	—	$100$
$1.00$	—	undefined
$1.01$	—	$-100^{\dagger}$
$1.03$	—	$-33.3^{\dagger}$
$2.00$	—	$-1^{\dagger}$
$^{\dagger}$ Past the wall these belong to a different solution; the one through $(0,1)$ ended at $x = 1$ .

💡 Every method in this lecture adds $h$ times an average of values of $y^2 \ge 0$ , so the computed values only climb: none of them can turn a positive $y$ negative, and the far branch below the axis stays out of reach. Refining the step makes the number reported at $x = 1$ larger, not better — $163.98$ at $h = 0.05$ , $819.91$ at $h = 0.01$ , $8199.10$ at $h = 0.001$ .

VQuiz Questions

Problem 1 · Locating the Wall

Given: $y' = y^2$ with $y(0) = 4$ — find the value of $x$ at which this solution blows up.

✅ Correct!

c = 1/y(0) = \tfrac14

, so the wall stands at

x = \tfrac14

and the solution exists only for

x < \tfrac14

❌ The relation is inverted. Substituting

x = 0

into

y = \frac{1}{c-x}

gives

y(0) = \frac{1}{c}

, so

c = 1/y(0)

, not

y(0)

❌ Not quite. The right-hand side really is smooth everywhere — that is exactly why the wall is invisible in the equation. The solution still ends at

x = c

❌ Not quite. Start from

y = \frac{1}{c-x}

, put in

x = 0

and

y = 4

, then ask where the denominator vanishes.

Show solution

Separating variables gives the family $y = \dfrac{1}{c-x}$ .

Step 1 — Find $c$ from the initial condition:

4 = y(0) = \frac{1}{c-0} = \frac{1}{c} \qquad\Longrightarrow\qquad c = \frac{1}{4}

Step 2 — Write the solution and find where the denominator vanishes:

y = \frac{1}{\tfrac14 - x}, \qquad \tfrac14 - x = 0 \;\Longrightarrow\; x = \tfrac14

Starting higher means blowing up sooner: $y(0) = 4$ reaches its wall at $x = \tfrac14$ , while $y(0) = 1$ survives until $x = 1$ . The interval of existence is $x < \tfrac14$ .

Problem 2 · Asking for a Value That Is Not There

Given: $y' = y^2$ with $y(0) = 1$ , so $y = \dfrac{1}{1-x}$ — what is $y(2)$ ?

✅ Correct! The solution through

(0,1)

lives only on

x < 1

. Asking it for

y(2)

asks for something it does not have.

❌ That is the far branch.

\frac{1}{1-2} = -1

is a genuine point of

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

, but on the piece with

x > 1

— a different solution of

y' = y^2

, not a continuation of this one across the wall.

❌ Not quite.

\infty

is what a numerical march prints, not a value the solution takes. The solution simply stops at

x = 1

❌ Not quite. Ask first whether

x = 2

lies inside the interval on which this solution exists.

Show solution

With $c = 1$ the solution is $y = \dfrac{1}{1-x}$ , and its denominator vanishes at $x = 1$ :

y(0.5) = 2, \qquad y(0.9) = 10, \qquad y(0.99) = 100, \qquad y(x) \to +\infty \text{ as } x \to 1^-

The formula $\frac{1}{1-x}$ still returns numbers for $x > 1$ , and those numbers do lie on a solution curve — the branch running from $-\infty$ up toward the axis. But an initial-value problem asks for the one curve through $(0,1)$ , and that curve is only the piece on $x < 1$ : integral curves cannot jump a singularity.

So the interval of existence is $-\infty < x < 1$ , and $y(2)$ is undefined for this solution.

Problem 3 · A Start Away from the Vertical Axis

Given: a solution of $y' = y^2$ passing through $\left(1, \tfrac12\right)$ — find $y(2)$ .

✅ Correct! Here

c = 3

, so

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

, the wall stands at

x = 3

, and

x = 2

is comfortably inside the interval of existence.

❌ That uses $c = 1/y_0$ , which only holds when the start is at $x_0 = 0$ . Substituting

x = 1

y = \tfrac12

into

y = \frac{1}{c-x}

gives

c - 1 = 2

, so

c = 3

and the wall is at

x = 3

, not

x = 2

❌ Not quite. Substitute the point into

y = \frac{1}{c-x}

to get

c

first, then evaluate the resulting formula at

x = 2

Show solution

Step 1 — Find $c$ from the point $\left(1, \tfrac12\right)$ :

\frac{1}{2} = \frac{1}{c-1} \qquad\Longrightarrow\qquad c - 1 = 2 \qquad\Longrightarrow\qquad c = 3

Step 2 — Write the solution:

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

Step 3 — Check that $x = 2$ is legal, then evaluate:

The wall sits at $x = c = 3$ , and the start $x = 1$ lies to its left, so this solution exists on $x < 3$ — which contains $x = 2$ .

y(2) = \frac{1}{3-2} = 1

Verify: $y' = \frac{1}{(3-x)^2} = \left(\frac{1}{3-x}\right)^2 = y^2$ ✓, and $y(1) = \frac{1}{2}$ ✓.

Problem 4 · Refining the Step

Given: RK4 on $y' = y^2$ from $(0,1)$ with $h = 0.01$ prints $819.91$ at $x = 1$ , where this solution has already ceased to exist — what does $h = 0.001$ print at $x = 1$ ?

✅ Correct! The printed value scales like

1/h

163.98

h = 0.05

819.91

h = 0.01

8199.10

h = 0.001

. A smaller step gives a bigger fiction.

❌ Wrong direction. The number grows like

1/h

, not like

h

: dividing the step by ten multiplies the printed value by about ten.

❌ It is a property of $h$ , not of the method. The same RK4 code prints

163.98

h = 0.05

and

8199.10

h = 0.001

❌ Convergence needs something to converge to. The solution through

(0,1)

has no value at

x = 1

at all, so no refinement of the step can approach one.

❌ Not quite. Compare the two reported values you were given, and ask how the number behaves as the step shrinks.

Show solution

The last true step is the one that reaches the wall. Near $x = 1$ the exact solution behaves like $y \approx \dfrac{1}{1-x}$ , so at the final grid point before the singularity the height is of size $1/h$ — and that is the size of whatever the march is holding when it arrives at $x = 1$ :

h = 0.05:\ 163.98 \qquad h = 0.01:\ 819.91 \qquad h = 0.001:\ 8199.10

Each tenfold refinement multiplies the printed value by about ten, so $h = 0.001$ prints roughly $8200$ .

Why the error orders do not apply. The statements "Euler's error $\approx Ch$ ", "improved Euler $\approx C_2h^2$ ", "RK4 $\approx C_4h^4$ " all compare the computed value with the value of the solution. Past $x = 1$ there is no such value, so there is nothing for the estimates to be estimates of. One step later the machine reports about $10^{13}$ , two more and it overflows to $\infty$ , and from then on every step adds $\infty$ to $\infty$ : at $x = 2$ the answer comes back $\infty$ .

The real lesson. Nothing in $y' = y^2$ mentions $c$ . The slope function is finite and smooth at every point of the plane, so no amount of slope-reading — at one point per step, or four — can announce in advance where the solution stops existing.

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The Wall the Equation Never Mentions: y′=y2y' = y^2y′=y2 and Its Private Singularities

IKey Formulas

IISeparating the Variables

IIIWhere the Wall Stands

IVMarching Into the Wall

VQuiz Questions

The Wall the Equation Never Mentions: $y' = y^2$ and Its Private Singularities