Differential-Equations · Unit 1 · Video 2 · Interactive Practice

Direction Fields and Isoclines: How to Draw an ODE by Hand

IKey Formulas

Formula	Name	What you need
$y' = f(x,y)$	Direction field — at $(x,y)$ hang a line element of slope $f(x,y)$	The equation in standard form
$y_1'(x) = f\big(x, y_1(x)\big)$	The theorem in one line: solution $\Longleftrightarrow$ integral curve	A candidate function $y_1$
$f(x,y) = C$	Isocline — the level curve of $f$ at level $C$	A slope $C$ , chosen before any point
$x - y^2 = C \iff x = y^2 + C$	Isoclines of the running example $y' = x - y^2$	The parabola $x = y^2$ shifted $C$ to the right

Key Insight: The isocline equation $f(x,y) = C$ carries no derivative — a differential equation cannot be plotted, only ordinary curves can. Those get dashed lines; solid lines are reserved for integral curves, which by the theorem are exactly the graphs of the solutions.

IIVisualization 1 — Pick the Slope First

Choosing the slope $C$ first collapses $y' = x - y^2$ into one ordinary curve, $x - y^2 = C$ .

Slope C 0

💡 Every element along $x = y^2 + C$ has slope $C$ , while the parabola's own slope is $\dfrac{1}{2y}$ — the two agree only at the single point $y = \dfrac{1}{2C}$ , and for $C = 0$ nowhere at all.

IIIVisualization 2 — Two Ways to Fill the Plane

Forty line elements: forty evaluations of $f$ point by point, or five dashed curves.

Order Elements drawn 40

IVVisualization 3 — Threading a Solution Through the Field

A curve tangent to every element it meets is exactly the graph of a solution.

x₀ 2 y₀ 1

💡 Two integral curves can never cross: at a crossing point the single line element there would have to supply two different slopes.

VQuiz Questions

Problem 1 · Reading One Line Element

Given: the direction field of $y' = x - y^2$ — find the slope of the line element at the point $(3, -2)$ .

✅ Correct!

3 - (-2)^2 = 3 - 4 = -1

, so the element at

(3,-2)

falls to the right, one unit down for one unit across.

❌ Not quite. That is

3 - (-2)

. The right-hand side squares

y

before subtracting:

(-2)^2 = 4

❌ Check the square. Squaring destroys the sign:

(-2)^2 = +4

, not

-4

, so you subtract 4 rather than adding it.

❌ Reversed. You computed

y^2 - x = 4 - 3

. The field's slope is

f(x,y) = x - y^2

, in that order.

❌ Not quite. Substitute

x = 3

and

y = -2

into

f(x,y) = x - y^2

Show solution

The direction field of $y' = f(x,y)$ hangs at each point $(x,y)$ a line element whose slope is the number $f(x,y)$ . Here $f(x,y) = x - y^2$ , so at $(3,-2)$ :

f(3,-2) = 3 - (-2)^2 = 3 - 4 = -1

The element has slope $\mathbf{-1}$ . Note that the answer depends on $y$ only through $y^2$ , so the elements at $(3,2)$ and $(3,-2)$ are identical — the whole field is symmetric across the $x$ -axis.

Problem 2 · The Isocline Equation

Given: $y' = x - y^2$ — find the isocline along which every line element has slope $C = -1$ .

✅ Correct!

x - y^2 = -1

rearranges to

x = y^2 - 1

: the parabola

x = y^2

shifted one unit left, drawn dashed, wearing elements of slope

-1

❌ Wrong side. On

x = y^2 + 1

the right-hand side is

x - y^2 = +1

, so those elements have slope

+1

, not

-1

❌ Check the algebra. Solving

x = y^2 - 1

for

y

gives

y = \pm\sqrt{x+1}

; a square root does not distribute across the

-1

❌ Wrong variable squared. The condition is

x - y^2 = C

, so

x

comes out as a function of

y

— the parabola opens to the right, not upward.

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

C

and solve the resulting ordinary equation:

x - y^2 = -1

Show solution

An isocline collects every point of the plane whose line element has one chosen slope $C$ . Those points satisfy $f(x,y) = C$ , with no derivative left in it:

x - y^2 = -1 \quad\Longrightarrow\quad x = y^2 - 1

That is the rightward-opening parabola $x = y^2$ pushed one unit to the left. Every point on it wears an element of slope $-1$ , and one plot of the curve serves all of them — that is the saving over evaluating $f$ point by point.

The family for a general $C$ is $x = y^2 + C$ : the same parabola sliding right as $C$ increases. Drawn dashed, because an isocline is scaffolding, not a solution.

Problem 3 · A New Field, Two Slopes

Given: $y' = y - x$ — find the isocline for $C = 2$ , and the slope of the line elements hanging on it.

Which curve is the isocline?

What slope do its line elements have?

✅ Correct! Along

y = x + 2

every element has slope 2, steeper than the line itself, so an integral curve reaching this isocline climbs across it to the upper side.

❌ Check the isocline. Set the right-hand side equal to the chosen slope:

y - x = 2

, then solve for

y

❌ That is the isocline's own slope. The line

y = x + 2

rises with slope 1, but the elements hung along it carry the slope you chose in advance.

❌ Not quite. The elements on the isocline

f(x,y) = C

all have slope

C

— that is the whole point of picking

C

first.

Show solution

Step 1 — the isocline. Set the right-hand side equal to $C = 2$ :

y - x = 2 \quad\Longrightarrow\quad y = x + 2

Step 2 — the elements. By construction every point of that line has $f(x,y) = 2$ , so each element there has slope $\mathbf{2}$ — not 1, which is the slope of the isocline as a line.

The general family: $y - x = C$ gives $y = x + C$ , a set of parallel lines of slope 1 whose elements get steeper as $C$ grows.

One member is special: for $C = 1$ the isocline $y = x + 1$ carries elements of slope 1 — the slope of the line itself. Substituting confirms it is a genuine solution: $y' = 1$ and $y - x = 1$ . That line is an isocline and an integral curve at once.

Problem 4 · Isocline and Integral Curve at Once

Given: $y' = 1 + x - y$ — exactly one of these lines is both an isocline of the field and an integral curve. Find it.

✅ Correct! On

y = x

the curve's own slope is

y' = 1

and the field's slope is

1 + x - x = 1

; the two agree, so this line is the

C = 1

isocline and a solution simultaneously.

❌ Not that one. Along

y = x + 1

the field gives

1 + x - (x+1) = 0

— horizontal elements crossing a line that climbs with slope 1.

❌ Not that one. Along

y = x - 1

the field gives

1 + x - (x-1) = 2

, so the elements are twice as steep as the line and cut across it.

❌ Not that one. Along

y = x + 2

the field gives

1 + x - (x+2) = -1

: elements pointing down across a line that rises.

❌ Not quite. A line is an integral curve only when its own slope equals the field's slope at every one of its points.

Show solution

Test the whole family $y = x + k$ at once. The graph's own slope is $y_1'(x) = 1$ . The field's slope along it is

f\big(x, y_1(x)\big) = 1 + x - (x + k) = 1 - k

The theorem says the graph is an integral curve exactly when those two agree:

1 = 1 - k \quad\Longrightarrow\quad k = 0

So $y = x$ is the only line of the family that solves the equation. It is also an isocline: the isoclines of $y' = 1+x-y$ are $y = x + 1 - C$ , and $k = 0$ corresponds to $C = 1$ — whose elements have slope 1, lying flat along the line and impossible to distinguish from it.

The other three lines are isoclines too ( $C = 0$ , $2$ and $-1$ respectively), but their elements cross them instead of lying along them, so none of them is a solution.

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