Differential-Equations · Unit 1 · Video 4 · Interactive Practice

Worked Example: $y' = 1 + x - y$ and the Isocline Corridor Trap

IKey Formulas

Formula	Name	What you need
$y' = 1 + x - y$	The equation	Not separable: $1 + x - y \neq f(x)\,g(y)$
$1 + x - y = C \;\Rightarrow\; y = x + 1 - C$	Isocline of slope $C$	One value of $C$
$y = x$ (the $C = 1$ isocline)	Isocline that is also a solution	Line slope $=$ carried slope
$x - 1 < y < x + 1 \iff 0 < y' < 2$	The corridor, $C = 0$ down to $C = 2$	Both walls

Key Insight: Every isocline is a line of slope $1$ , but the slope it carries is $C$ — so exactly one member of the family, $C = 1$ , moves the way it tells everything else to move.

IIVisualization 1 — The Isocline Family

Each isocline climbs at slope $1$ ; the slope it hands out is $C$ . When do the two agree?

Element slope C 0

💡 Separation of variables never starts here: the sum $1 + x - y$ cannot be written as $f(x)\,g(y)$ , so the only formula-producing method met so far fails before the first integral.

IIIVisualization 2 — Why Both Doors Are Locked

Promote the $C = 0$ and $C = 2$ isoclines to walls: both aim their elements into the corridor.

Step 1

Step 1 — The corridor

\text{upper wall } (C = 0):\; y = x + 1 \qquad \text{lower wall } (C = 2):\; y = x - 1

x - 1 < y < x + 1 \iff 0 < 1 + x - y < 2

Strictly inside, every slope is positive and none reaches

2

; the solution

y = x

runs up the centre.

Step 2 — The upper door

To cross

y = x + 1

going up, a curve must out-climb the wall: it needs

y' > 1

there. The equation decides instead:

\text{on } y = x+1: \quad y' = 1 + x - (x+1) = 0

The curve arrives level while the wall climbs at slope

1

, so the wall pulls away above it and the curve drops back inside.

Step 3 — No doubling back

The only other route over the top is to turn back, and turning back needs

y' < 0

somewhere inside. But inside the corridor

0 < y' < 2,

so no interior point offers a negative slope. ❌ The upper door is locked.

Step 4 — The lower door

To drop through

y = x - 1

, a curve must fall behind the wall: it needs

y' < 1

there. Again the equation decides:

\text{on } y = x-1: \quad y' = 1 + x - (x-1) = 2

Slope

2

against a wall of slope

1

— the elements spike back into the corridor. ❌ The lower door is locked too.

Step 5 — Trapped for good

|y(x_0) - x_0| < 1 \;\Longrightarrow\; |y(x) - x| < 1 \quad \text{for every } x > x_0

A curve that enters climbs on inside forever, held within distance

1

of the line

y = x

— settled without solving anything.

IVVisualization 3 — Release a Solution

Can a curve released outside the corridor miss it, or one released inside ever escape?

💡 The trapped curve draws closer and closer to $y = x$ without ever appearing to meet it — whether two integral curves can cross at all is where the geometry goes next.

VQuiz Questions

Problem 1 · Read Off an Isocline

Given: $y' = 1 + x - y$ — find the isocline along which every line element has slope $C = 3$ .

✅ Correct! On

y = x - 2

the field gives

1 + x - (x - 2) = 3

at every point — a line of slope

1

carrying elements of slope

3

❌ Not quite.

C

is the slope of the elements, not of the isocline. Solving

1 + x - y = C

always leaves the coefficient of

x

equal to

1

, so every isocline in this family has slope

1

❌ Close, but check the sign. Solving

1 + x - y = C

for

y

gives

y = x + 1 - C

, so a larger

C

pushes the line down, not up.

❌ Close. The constant is

1 - C

, not

-C

— the

1

on the right-hand side of

y' = 1 + x - y

has to travel too.

❌ Not quite. Set the right-hand side equal to

3

and solve for

y

Show solution

An isocline is the set of points where the field prescribes one fixed slope, so set the right-hand side equal to $C = 3$ :

1 + x - y = 3

Solve for $y$ in one step:

y = x + 1 - 3 = x - 2

Check: at any point of $y = x - 2$ the equation gives $y' = 1 + x - (x - 2) = 3\ \checkmark$

Note the two features the general formula $y = x + 1 - C$ shows: the coefficient of $x$ is $1$ for every $C$ (the whole family is parallel, slope $1$ ), and raising $C$ lowers the line.

Problem 2 · The Line That Is Both

Given: $y' = 1 + x - y$ — exactly one isocline of this equation is also an integral curve. Which line is it?

✅ Correct! On

y = x

the field prescribes slope

1 + x - x = 1

, and the line delivers slope

1

— an exact solution read straight off the picture.

❌ Not quite.

y = x + 1

is the

C = 0

isocline: it carries horizontal elements while climbing at slope

1

, so its own direction disagrees with the field on it.

❌ Not quite.

y = x - 1

is the

C = 2

isocline: its elements climb at slope

2

, twice the line's own slope.

❌ Not quite. Every isocline here has slope

1

and carries slope

C

; the two agree only for one value of

C

Show solution

An integral curve moves, at every point, with exactly the slope the field prescribes there. So an isocline is also an integral curve precisely when its own slope equals the slope it carries:

\underbrace{1}_{\text{slope of every isocline}} = \underbrace{C}_{\text{slope it carries}} \;\Longrightarrow\; C = 1

Put $C = 1$ into $y = x + 1 - C$ :

y = x + 1 - 1 = x

Verify by substitution. If $y = x$ then the left side is $y' = 1$ , and the right side is

1 + x - y = 1 + x - x = 1

Both sides equal $1$ everywhere along the line, so $y = x$ is an exact solution — obtained with no integration at all. The coincidence is unique because the family has slope $1$ for every $C$ , so only $C = 1$ can match.

Problem 3 · Inside the Corridor

Given: $y' = 1 + x - y$ and the corridor between the walls $y = x + 1$ ( $C = 0$ ) and $y = x - 1$ ( $C = 2$ ).

What slopes occur strictly inside the corridor?

A solution reaches the upper wall at the point $(3, 4)$ . What slope does the equation assign it there?

✅ Correct! Interior slopes fill

(0, 2)

, and on the upper wall the field flattens the curve to slope

0

while the wall keeps climbing at slope

1

— so the curve cannot follow it out.

❌ Check the endpoints. The walls are the

C = 0

and

C = 2

isoclines, and

y' = 1 + x - y

moves continuously between those two values as

y

sweeps across the band.

❌ That is the wall's own slope, not the curve's. At the moment of contact the differential equation, not the wall, dictates the curve's slope: substitute

x = 3

y = 4

into

1 + x - y

❌ Not quite. Substitute the point straight into the right-hand side:

1 + 3 - 4

Show solution

Part 1 — the interior slopes. A point is strictly inside the corridor exactly when

x - 1 < y < x + 1

Take the two halves separately and move $y$ across; subtracting from $1 + x$ flips each inequality:

y < x + 1 \;\Longrightarrow\; 1 + x - y > 0, \qquad y > x - 1 \;\Longrightarrow\; 1 + x - y < 2

So $0 < y' < 2$ everywhere inside: every interior slope is positive, which is exactly what rules out doubling back over the top.

Part 2 — the slope at $(3, 4)$ . The point is on the upper wall, since $4 = 3 + 1$ . Substitute into the equation:

y' = 1 + x - y = 1 + 3 - 4 = 0

The curve arrives horizontally while the wall climbs at slope $1$ . The wall outruns it, and the curve falls back inside the corridor. The same substitution on the lower wall gives $y' = 1 + x - (x - 1) = 2$ , steeper than the wall — pointing back in as well.

Problem 4 · The Same Trap, a New Equation

Given: the different equation $y' = x - y$ . Its isoclines $x - y = C$ are again parallel lines of slope $1$ .

Which isocline of $y' = x - y$ is also an integral curve?

Which two isoclines are its trapping walls — upper wall carrying slope $0$ , lower wall carrying slope $2$ ?

✅ Correct! The whole picture just slides down by

1

: the solution is

y = x - 1

, and its corridor runs between

y = x

and

y = x - 2

❌ That was the answer to the old equation. Test it: on

y = x

the new field gives

y' = x - x = 0

, but the line climbs at slope

1

— so the line and the field disagree.

❌ Not quite. Write the family as

y = x - C

: each member has slope

1

and carries slope

C

, so match them.

❌ Not quite. Solve

x - y = 0

for the upper wall and

x - y = 2

for the lower wall.

Show solution

The family. Setting $x - y = C$ gives

y = x - C

so once again every isocline has slope $1$ and the member indexed by $C$ carries slope $C$ .

Part 1 — the self-consistent line. Slope of the line equals slope carried when $1 = C$ , i.e. $C = 1$ :

y = x - 1

Verify: $y' = 1$ and $x - y = x - (x - 1) = 1\ \checkmark$

Part 2 — the walls. The upper wall is the isocline carrying slope $0$ and the lower wall the isocline carrying slope $2$ :

C = 0:\; y = x, \qquad C = 2:\; y = x - 2

Strictly between them $x - 2 < y < x$ , hence $0 < x - y < 2$ : all interior slopes are positive. On $y = x$ the field gives slope $0$ against a wall of slope $1$ , and on $y = x - 2$ it gives slope $2$ against the same slope $1$ — both doors locked, with the solution $y = x - 1$ running up the centre.

Solved: 0 / 4

Worked Example: y′=1+x−yy' = 1 + x - yy′=1+x−y and the Isocline Corridor Trap

IKey Formulas

IIVisualization 1 — The Isocline Family

IIIVisualization 2 — Why Both Doors Are Locked

IVVisualization 3 — Release a Solution

VQuiz Questions

Worked Example: $y' = 1 + x - y$ and the Isocline Corridor Trap