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

First-Order ODEs: Why Most Can't Be Solved, and the Geometric View

IKey Formulas

Formula	Name	What you need
$\dfrac{dy}{dx} = f(x, y)$	Standard form of a first-order ODE	The derivative alone on the left
$\text{slope at } (x, y) = f(x, y)$	Line element — the direction field	A point and the equation
$y_1'(x) = f\big(x,\, y_1(x)\big)$	Integral curve (solution) condition	Tangency at every point of the curve
$y' = \dfrac{x}{y} \;\Longrightarrow\; y\,dy = x\,dx$	Separable — the one of the three you can already solve	Variables that come apart

Key Insight: The dictionary has exactly two entries: the equation $y' = f(x,y)$ becomes a direction field, and a solution $y_1(x)$ becomes an integral curve. Both are read off $f$ point by point, so the picture still exists for $y' = x - y^2$ , whose solutions no elementary formula can write down.

IIVisualization 1 — The Equation Prescribes a Slope

Where $f$ is defined, the equation hands out one number: the slope of the line element planted there.

Equation

IIIVisualization 2 — Tangent at Every Point

An integral curve matches the field's slope at every point it passes, not merely at one.

Curve x 0

💡 A curve that slips between two drawn elements misses nothing: the plane can never be filled in completely, and tangency is required at every point of the curve — drawn or undrawn.

IVVisualization 3 — A Curve Built Without a Formula

$y' = x - y^2$ has no elementary solution, yet the curve through each point is fully determined.

💡 The tame twin $y' = y - x^2$ paints a field of exactly the same kind; a solution formula is a convenience the picture never needs.

VQuiz Questions

Problem 1 · Reading a Line Element

Given: the equation $y' = x - y^2$ — find the slope of the line element planted at the point $(1, 3)$ .

✅ Correct!

f(1,3) = 1 - 3^2 = 1 - 9 = -8

— a steep downhill dash, read straight off the equation.

❌ Not quite. The

y

term is squared before the subtraction:

3^2 = 9

, not

3

❌ Check the order. The right-hand side is

x - y^2

, not

y^2 - x

❌ Not quite. Only

y

is squared — the subtraction happens after, not inside, the square.

❌ Not quite. Substitute

x = 1

and

y = 3

into

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

Show solution

The equation is already in standard form, so $f(x, y) = x - y^2$ , and the line element at a point carries the slope $f$ takes there:

f(1, 3) = 1 - 3^2 = 1 - 9 = -8

No equation was solved to get this. The slope comes from substitution alone, which is why a direction field can be painted even for an equation with no solution formula.

Problem 2 · Testing a Candidate Curve

Given: the equation $y' = x/y$ and the curve $y = 1 + \tfrac{1}{2}x^2$ . At $(0, 1)$ the curve runs horizontally and the field prescribes $\tfrac{0}{1} = 0$ — the two agree. Is the curve an integral curve of $y' = x/y$ ?

✅ Correct! One match at

(0,1)

proves nothing: at

(2,3)

the curve climbs three times too steeply, so it cuts across the field.

❌ Close, but the definition is stronger. Tangency must hold at every point of the curve, not at one of them.

❌ Not quite. That the field is defined along the curve says nothing about whether the curve matches it.

❌ Not quite. Tangency is checked point by point from

f(x,y)

alone — no solution formula is needed.

❌ Not quite. Compare

y'

of the curve with

x/y

at a second point, say

x = 2

Show solution

Tangency is required at every point of the curve, so test a second point. At $x = 2$ the curve is at $y = 1 + \tfrac{1}{2}(2)^2 = 3$ , and

\text{curve slope} = y' = x = 2, \qquad \text{field slope} = \frac{x}{y} = \frac{2}{3}.

The two disagree, so the curve crosses the line element at $(2,3)$ instead of running along it: not an integral curve.

Separating the variables gives $y\,dy = x\,dx$ , hence $y^2 - x^2 = C$ — the integral curves of this field are hyperbolas, and the one through $(0,1)$ is $y = \sqrt{1 + x^2}$ , not a parabola.

Problem 3 · Verifying a Solution of the Tame Twin

Given: the equation $y' = y - x^2$ and the candidate $y_1(x) = x^2 + 2x + 2$ — check the two sides of the equation separately.

What is $y_1'(x)$ ?

What is $y_1(x) - x^2$ ?

✅ Correct! Both sides equal

2x + 2

for every

x

, so

y_1

satisfies

y' = y - x^2

and its graph is an integral curve of that field.

❌ Check the derivative. Differentiate term by term:

x^2 \to 2x

\;2x \to 2

\;2 \to 0

❌ Check the subtraction. Only the

x^2

term is cancelled; the rest of

y_1

survives unchanged.

Show solution

Step 1 — differentiate the candidate.

y_1(x) = x^2 + 2x + 2 \quad \Longrightarrow \quad y_1'(x) = 2x + 2

Step 2 — evaluate the right-hand side along the candidate.

y_1(x) - x^2 = (x^2 + 2x + 2) - x^2 = 2x + 2

Step 3 — compare. The two sides agree for every $x$ , so $y_1$ is a solution of $y' = y - x^2$ — the tame twin from the video.

Geometrically the parabola $y = x^2 + 2x + 2$ is an integral curve of that field. At $(0, 2)$ , for instance, the field prescribes $2 - 0^2 = 2$ and the curve's own slope is $2(0) + 2 = 2$ .

Problem 4 · What "Unsolvable" Means

Given: no elementary function solves $y' = x - y^2$ . A student concludes that nothing at all can be said about the solution through the origin. Which statement is correct?

✅ Correct!

f(0,0) = 0 - 0^2 = 0

. "Unsolvable" limits the reach of our notation, not the existence of the curves.

❌ Not quite. A slope of

0

is a perfectly good line element — a horizontal one.

❌ Not quite. The solutions exist as smooth curves through the whole plane; only a formula for them is missing.

❌ Not quite. The curve is built by following the field, slope after slope — the geometric view never consults a formula.

❌ Not quite. Evaluate

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

at the origin before deciding what is known.

Show solution

"Unsolvable" is a statement about formulas, not about existence. The right-hand side $f(x, y) = x - y^2$ can be evaluated at any point whatsoever, so the direction field is defined everywhere. At the origin

f(0, 0) = 0 - 0^2 = 0,

so the integral curve through $(0,0)$ crosses it horizontally, and following the field from there fixes the rest of the curve, point after point.

What fails is only the notation: no finite combination of powers, roots, exponentials, logarithms, sines and cosines writes this solution down. The curve is there regardless.

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