Differential-Equations · Unit 5 · Video 1 · Interactive Practice

No t on the Right: Autonomous Equations and the Direction Field That Never Changes

IKey Formulas

Formula	Name	What it says
$\dfrac{dy}{dt} = f(y)$	Autonomous equation	No $t$ on the right: the slope depends only on the current value of $y$
$y = y_0 \;\Longrightarrow\; \dfrac{dy}{dt} = f(y_0)$	Horizontal isoclines	Every horizontal line is an isocline, carrying the single slope $f(y_0)$
$\displaystyle\int \frac{dy}{f(y)} = t + c$	Always separable	Gives $t$ in terms of $y$ ; getting $y(t)$ still means inverting it
$u(t) = y(t - t_0)$	Time translate	If $y$ is a solution, so is $u$ : the same graph pushed $t_0$ units to the right

Key Insight: With no $t$ on the right, the direction field is the same at every time. Its isoclines are the horizontal lines, and any integral curve slid sideways is again an integral curve: the shape of a solution depends on where it starts, not on when.

IIEvery Horizontal Line Is an Isocline

Without $t$ on the right, the slope is constant along every horizontal line; with $t$ , it changes along the line.

Equation Probe t 1.5 Probe y 0.5

IIISlide a Solution Sideways

A solution of $dy/dt = 1 - y$ slid $t_0$ units right is still a solution; one of $dy/dt = 1 + t - y$ is not.

Equation Shift t₀ 1.5

💡 The separated solution says the same thing: with $F' = 1/f$ , $\int dy/f(y) = t + c$ reads $t = F(y) - c$ , and changing $c$ only slides that graph sideways.

IVQuiz Questions

Problem 1 · Which One Is Autonomous?

Given: four first-order equations — find the one that is autonomous.

✅ Correct! Here the unknown is

x

and the independent variable is

t

, and the right-hand side

x^2 - 1

is built from

x

alone.

❌ Not quite. This right-hand side is a function of one variable, but it is the independent variable

x

: the slope is constant along vertical lines, not horizontal ones.

❌ Not quite. One letter away from

y^3 - y

: the

t

on the right makes the slope at a given height change with time.

❌ Not quite. The factor

t

multiplies the whole right-hand side, so at height

y = 2

the slope is

3t

— it depends on when you are there.

Show solution

An equation is autonomous when its right-hand side is a function of the dependent variable (the unknown) alone.

(a) The unknown is $y$ and the independent variable is $x$ ; the right-hand side $x^2 - 1$ contains only $x$ . Not autonomous — it is a pure integration, $y = \tfrac{x^3}{3} - x + c$ .
(b) $y^3 - t$ contains the independent variable $t$ . Not autonomous.
(c) The unknown is $x$ , the independent variable is $t$ , and $x^2 - 1$ contains only $x$ . Autonomous.
(d) $t\,(y^2 - 1)$ contains $t$ ; at $y = 2$ the slope is $3t$ . Not autonomous.

The same expression $x^2 - 1$ is autonomous in (c) and not in (a): what decides it is which variable is the unknown.

Problem 2 · Which Way Is the Shift?

Given: $y(t)$ is a non-constant solution of $\dfrac{dy}{dt} = f(y)$ , defined for all $t$ , with $y(0) = 2$ — find the function that is guaranteed, whatever $f$ is, to be a solution with value $2$ at $t = 3$ .

✅ Correct!

u(3) = y(0) = 2

, and

u'(t) = y'(t - 3) = f\big(y(t - 3)\big) = f\big(u(t)\big)

❌ Check the direction.

y(t + 3)

is a solution, but it takes the value

y(0) = 2

t = -3

: replacing

t

t + 3

pushes the graph 3 units to the left.

❌ Not quite.

u(3) = 2

, but the inner function

3 - t

has derivative

-1

, so

u'(t) = -f\big(u(t)\big)

u

solves

du/dt = -f(u)

, with time running backward.

❌ Not quite.

u(3) = 2

, but moving the graph vertically changes its height and not its slopes:

u'(t) = f\big(y(t)\big)

, while the field at height

u(t)

asks for

f\big(u(t)\big)

. Only horizontal slides are guaranteed.

Show solution

Take $u(t) = y(t - 3)$ . Then $u(3) = y(0) = 2$ , and by the chain rule, with inner derivative $1$ ,

u'(t) = y'(t - 3) = f\big(y(t - 3)\big) = f\big(u(t)\big).

So $u$ solves the same equation; it is the graph of $y$ pushed 3 units to the right.

$y(t + 3)$ is also a solution, but it is the graph pushed 3 units to the left: it equals $2$ at $t = -3$ , while at $t = 3$ it equals $y(6)$ , not $y(0)$ .
$y(3 - t)$ : $u'(t) = -y'(3 - t) = -f\big(u(t)\big)$ , the equation with time reversed.
$y(t) + 2 - y(3)$ : $u'(t) = f\big(y(t)\big)$ is the slope belonging to the old height, but $u$ sits at a different height, where the field demands $f\big(u(t)\big)$ .

Check with $f(y) = 1 - y$ , whose solution with $y(0) = 2$ is $y = 1 + e^{-t}$ : only $y(t - 3) = 1 + e^{3 - t}$ satisfies both $u' = 1 - u$ and $u(3) = 2$ .

Problem 3 · Shifting a Non-Autonomous Solution

Given: $y(t)$ solves $\dfrac{dy}{dt} = 1 + t - y$ — find the equation satisfied by the shifted function $u(t) = y(t - 2)$ .

✅ Correct!

u'(t) = y'(t - 2) = 1 + (t - 2) - u(t) = t - 1 - u(t)

— the shift has changed the equation.

❌ Not quite. Keeping the equation unchanged is exactly the step that needs autonomy. The chain rule evaluates the right-hand side at time

t - 2

, so its

t

becomes

t - 2

❌ Check the sign. The inner argument is

t - 2

, so the

t

on the right becomes

t - 2

, not

t + 2

❌ Check the chain rule. The derivative of the inside,

t - 2

, is

+1

: there is no sign flip.

Show solution

By the chain rule, $u'(t) = y'(t - 2) \cdot 1$ . Because $y$ solves the equation at every time $s$ , $y'(s) = 1 + s - y(s)$ ; apply this at $s = t - 2$ :

u'(t) = 1 + (t - 2) - y(t - 2) = t - 1 - u(t).

So $\dfrac{du}{dt} = t - 1 - u$ : the $t$ on the right has been replaced by $t - 2$ , which is the $f(t - t_0, u)$ of the video — a different equation.

Check with the solution $y = t$ (it satisfies $y' = 1 = 1 + t - t$ ): the shift $u = t - 2$ has $u' = 1$ and $t - 1 - (t - 2) = 1$ ✓, whereas $1 + t - u = 3 \neq 1$ .

Problem 4 · The Mirror Image of Autonomy

Given: the right-hand side of $\dfrac{dy}{dx} = x^2 - 1$ depends on the independent variable alone — find the true statement about its isoclines and its solutions.

✅ Correct! The slope

x_0^2 - 1

is the same at every point of the vertical line

x = x_0

, and the solutions

y = \tfrac{x^3}{3} - x + c

differ only by the vertical shift

c

❌ Not quite. That is the autonomous picture. Here the slope depends on

x

alone, so it changes as you move sideways and stays fixed as you move up or down.

❌ Half right. The isoclines are vertical, but a horizontal slide changes

x

\tfrac{(x - h)^3}{3} - (x - h)

has slope

(x - h)^2 - 1

, not

x^2 - 1

❌ Half right. Vertical slides do work, but along a horizontal line

y = y_0

the slope

x^2 - 1

still changes with

x

; it is constant along vertical lines.

Show solution

The slope at $(x, y)$ is $x^2 - 1$ , whatever $y$ is. Along the vertical line $x = x_0$ it is the single number $x_0^2 - 1$ , so every vertical line is an isocline.

Integrating directly, $y = \tfrac{x^3}{3} - x + c$ : every solution is the graph of $\tfrac{x^3}{3} - x$ pushed up or down by $c$ , so a vertical slide of a solution is a solution. A horizontal slide by $h \neq 0$ is not:

\frac{d}{dx}\left[\frac{(x - h)^3}{3} - (x - h)\right] = (x - h)^2 - 1 \neq x^2 - 1.

The two special cases swap roles: $dy/dt = f(y)$ has horizontal isoclines and horizontal translates; $dy/dx = g(x)$ has vertical isoclines and vertical translates.

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