Learn the language of change

A visual, interactive workbook

Differential equations stop feeling like a bag of tricks when you can read what an equation says about motion. This book trains that reading skill first, then turns it into reliable solution habits.

Start with state, rate, and solution Jump to mixed practice

What you will learn to do

Most courses organize differential equations by chapter titles. Problems rarely arrive with their chapter title attached. This workbook organizes your attention around four questions:

1What changes?

Name the state and its units.

2What sets the rate?

Read signs, factors, and feedback.

3Which fingerprint fits?

Choose a method from structure.

4Does the answer behave?

Check units, shape, and substitution.

By the end, you should be able to move among four representations of the same model:

ASituation

→

BEquation

→

CField

→

DSolution

↺

The workout rhythm

Each chapter repeats the same deliberate sequence.

Notice. Predict from a picture, a story, or the equation's shape.
Name. Learn one small piece of mathematical language.
Work. Study a fully explained example.
Fade. Complete a similar example with fewer steps supplied.
Choose. Mix the new method with older ones so that you must identify the method yourself.
Check. Test the result against the original equation and its expected behavior.

Your practice contract

Make a prediction before touching the algebra. Write one line for every transformation. Open a solution only after committing to an attempt. When an answer is wrong, name the first line where the reasoning changed course.

What you need

You should be comfortable with derivatives, antiderivatives, exponentials, logarithms, and basic algebra. Chapter 9 introduces the small amount of linear algebra it needs. A calculator helps with arithmetic, but every central idea works without special software.

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A two-minute preflight

Try these without notes. The point is to choose a starting pace, not to earn a score.

1. If $y'(t)\lt 0$ on an interval, what must the graph of $y$ do there?

2. Which derivative rule recognizes $e^{2t}y' + 2e^{2t}y$ ?

If either question felt unfamiliar, work straight through Parts I and II. If both felt immediate, skim Chapter 1 and spend more time on the integrated problems.

I made an honest preflight attempt.

Next: Learn to separate the state, the rate, and the solution.

State, rate, and solution

Chapter 1 - Read before you solve

An ordinary differential equation is a sentence about change. It names a state, gives a rule for the state's rate, and asks for a solution function whose motion obeys that rule.

By the end of this chapter, you should be able to:

identify the independent variable, state, rate, and parameters;
translate a verbal rate law into an equation;
distinguish a differential equation from one of its solutions; and
verify a proposed solution by substitution and by behavior.

The three nouns

State

y(t)

what can change

→

Rate

y'(t)

how it changes now

→

Rule

y'=f(t,y)

what controls the rate

→

Solution

y(t)

a whole history

↺

Suppose $T(t)$ measures a cup of coffee's temperature in degrees Celsius, while the independent variable $t$ measures minutes. Then:

$T$ is the state, with units °C;
$T'$ is the rate, with units °C/min;
room temperature, say $22$ °C, and a cooling constant $k$ are parameters, fixed quantities that set the model's behavior;
$T'=-k(T-22)$ is the differential equation; and
$T(t)=22+68e^{-kt}$ is the solution that starts at $90$ °C.

The equation is a local rule: it tells the slope at the current state. A solution collects those local instructions into an entire curve.

Say the equation aloud

"The temperature changes at a rate proportional to its signed distance from room temperature, directed back toward the room." If you cannot say an equation in words, the symbols have not become usable language yet.

Units expose meaning

Terms added together must have the same units. In

$T'=-k(T-22),$

the left side has units °C/min. The factor $(T-22)$ has units °C, so $k$ must have units $1/\text{min}$ .

This gives a fast error detector. The equation $T'=-(T-22)^2$ cannot use a proportionality constant with no units: its right side would have units °C² rather than °C/min.

Unit check. In $P'=rP(1-P/K)$ , population $P$ is measured in cells and time in hours. What are the units of $r$ ?

Worked example: read a model from a story

Medicine leaves the bloodstreamfully worked

A patient starts with 120 mg of a medicine in the bloodstream. The body removes 18% of the current amount per hour. Build an initial-value problem, an equation paired with a starting value, and predict the graph before solving.

Name the state. Let $A(t)$ be the amount in milligrams after $t$ hours. Then $A'$ has units mg/hour.
Translate the rate phrase. "18% of the current amount per hour" gives magnitude $0.18A$ . "Removes" supplies the negative sign: $A'=-0.18A$ .
Attach the initial condition. "Starts with 120 mg" means $A(0)=120$ .
Predict before solving. When $A\gt 0$ , the rate is negative, so the graph decreases. As $A$ shrinks, $|A'|$ shrinks, so the curve flattens toward zero.
State the model. $A'=-0.18A,\quad A(0)=120$ . Chapter 4 will solve it; for now, its behavior already says a great deal.

Fade the support

I do

"Grows by 7% of its current size per year" becomes $P'=0.07P$ .

We do

"Loses 4 grams plus 3% of its current mass per day" becomes $M'=\underline{\hspace{3em}}$ .

Check the two rate contributions

$M'=-4-0.03M$ .

You do

A tank receives 5 L/min of clean water and drains 5 L/min. If $S$ is grams of salt in a 100 L tank, write the salt rate.

One cue

Rate in minus rate out. Clean inflow carries no salt.

The last model is $S'=0-(5\text{ L/min})(S/100\text{ g/L})=-S/20$ g/min.

A solution is a function, not a final number

Students often carry an algebra habit into differential equations: "solve" should produce a number. Here it usually produces a function.

For $y'=2y$ :

the differential equation names a family of possible motions;
$y=Ce^{2t}$ is the general solution family;
$y(0)=3$ selects the particular solution $y=3e^{2t}$ ; and
$y(4)=3e^8$ is one value of that solution.

One rate law can admit many solution curves. An initial condition chooses one.

Verify in three passes

To test the claim that $y=4e^{-3t}$ solves $y'=-3y$ with $y(0)=4$ :

Differentiate: $y'=-12e^{-3t}$ .
Substitute: $-3y=-3(4e^{-3t})=-12e^{-3t}=y'$ .
Check the condition and behavior: $y(0)=4$ , the curve stays positive, decreases, and flattens. That matches a negative rate whose magnitude shrinks with $y$ .

Symbolic substitution proves the equation. Behavior checks catch sign and interpretation errors that algebra can hide.

Error clinic: checking only the initial value

A student claims $y=4-3t$ solves $y'=-3y$ because $y(0)=4$ . The condition is necessary, but it is not enough. Here $y'=-3$ , while $-3y=-12+9t$ ; they disagree except at one accidental time.

Practice

Practice - translation
1.1 A pond contains $W(t)$ cubic meters of water. Rain adds 12 m³/hour, while leakage removes 4% of the current volume per hour. Write the differential equation and state the units of every term.
Detailed solution
Practice - verification
1.2 Verify whether $y=2+3e^{-t}$ solves $y'=2-y$ and $y(0)=5$ . Use all three passes.
Detailed solution
Integrated - representation
1.3 For $N'=0.4N(1-N/50)$ , decide whether the state initially rises or falls when $N(0)=20$ , $50$ , and $80$ . Do not solve.
Detailed solution
Challenge - modeling
1.4 A 200 L tank initially holds 10 kg of dissolved salt. Brine containing 0.08 kg/L enters at 3 L/min; well-mixed solution leaves at 3 L/min. Let $S(t)$ be kilograms of salt. Build the initial-value problem and identify the long-run amount suggested by a zero rate.
Detailed solution
Practice - extension
1.5 A dye amount $D(t)$ , measured in grams, obeys $D'=-0.2D$ , with $t$ in hours and $D(0)=10$ . Find the initial rate, verify $D=10e^{-0.2t}$ , and compare its one-hour percentage loss with an exact 20% loss each hour.
Detailed solution
Challenge - transfer
1.6 A well-mixed tank initially contains 100 L of clean water. Brine with concentration 3 g/L enters at 2 L/min, while solution leaves at 1 L/min. Assuming no overflow, write $V(t)$ , the initial-value problem for salt amount $S(t)$ , and the initial salt rate. Explain why the outflow concentration changes with time.
Detailed solution

Retrieve before you leave

Close the page or cover the text. Say these answers aloud:

What are the four objects you name when reading a model?
Why does a differential equation usually have a family of solutions?
What three passes verify a proposed solution?

I answered the retrieval questions without looking, then corrected what I missed.

Next: See the field of slopes before choosing any solution method.

See the field before solving

Chapter 2 - Turn a rate rule into a picture

An equation $y'=f(t,y)$ assigns one slope to every point $(t,y)$ . A direction field draws a small segment with that slope at each point. A solution curve does not invent its path; it threads through the assigned segments.

By the end of this chapter, you should be able to:

read increasing, decreasing, and equilibrium behavior from a field;
sketch solution curves through initial conditions;
connect features of $f(t,y)$ to visible features of the field; and
identify what a numerical or symbolic answer must look like before calculating it.

Lab: move the initial condition

Direction-field explorerEvery short mark is a local instruction.

Rate rule

local slopesolution curveinitial condition

Try these moves before reading on:

Choose $y'=-0.65y$ . Place the initial point above, on, and below $y=0$ .
Choose the logistic equation. Find every horizontal row where the marks go flat.
Choose $y'=\sin t-0.35y$ . Notice that equal heights can have different slopes at different times.

Read a field in four passes

1. Find zero-slope sets

Where $f(t,y)=0$ , the field is horizontal. If the equation is autonomous, $y'=f(y)$ , its rate depends only on the state, and a horizontal zero-slope row often gives an equilibrium solution: a state that stays constant.

For $y'=y(4-y)$ , the zero-rate states are $y=0$ and $y=4$ .

2. Mark the sign regions

$y'\gt 0$ : solution curves rise as $t$ increases.
$y'\lt 0$ : solution curves fall as $t$ increases.
large $|y'|$ : solution curves look steep.

For $y'=y(4-y)$ , the rate is positive between 0 and 4 and negative above 4. Positive solutions move toward 4.

3. Ask what stays constant

If $f$ depends only on $y$ , every mark on one horizontal row has the same slope. If $f$ depends only on $t$ , every mark in one vertical column has the same slope. These repeated patterns identify structure before algebra does.

4. Thread, do not connect dots

A solution curve stays tangent to the field and remains smooth where $f$ is smooth. It cannot turn at a point where the assigned slope is strictly positive or cross an equilibrium when uniqueness applies.

Field fingerprint. A direction field looks identical along every horizontal row but changes from one row to the next. What form is most likely?

Worked example: predict without solving

Read $y'=2-y$ , $y(0)=5$ fully worked

Zero rate: $2-y=0$ at $y=2$ . The row $y=2$ is horizontal.
Sign: above $y=2$ , $y'\lt 0$ ; below it, $y'\gt 0$ . Arrows point toward 2 from both sides.
Initial motion: $y'(0)=2-5=-3$ , so the curve starts downward with slope $-3$ .
Long-run prediction: the rate weakens as $y$ nears 2, so the curve falls and flattens toward $y=2$ .
Shape check: differentiating the equation gives $y''=-y'$ . While the solution falls, $y'\lt 0$ , so $y''\gt 0$ . The curve is concave up.

Same structure, less support

For $y'=1-\tfrac12 y$ , $y(0)=-2$ :

the equilibrium is $y=\underline{\hspace{2em}}$ ;
the initial slope is $\underline{\hspace{3em}}$ ;
the solution moves [toward / away from] the equilibrium; and
while it rises, its concavity is [up / down].

Check the four predictions

The equilibrium is $y=2$ ; the initial slope is $2$ ; the solution moves toward 2; and $y''=-\tfrac12 y'\lt 0$ , so it is concave down while rising.

What can go wrong?

Error clinic: treating a slope mark as a point to hit

The small segment at $(1,2)$ reports the slope a solution would have if it passed through $(1,2)$ . It does not require every solution to pass through that point. A field encodes infinitely many possible initial-value problems at once.

Error clinic: drawing vertical solution segments

A classical solution $y(t)$ is a function of $t$ , so its graph cannot double back or become vertical. Very steep is possible; vertical is not a finite derivative.

When may solution curves cross?

If $f(t,y)$ and its dependence on $y$ are sufficiently well behaved near a point, one initial condition selects one local solution. Two distinct solution curves cannot cross there: the crossing point would give the same initial condition two different futures.

This is a conceptual use of uniqueness. You do not need the full theorem yet. Use the practical rule: for the smooth equations in the early chapters, never sketch two solution curves crossing.

Practice

Practice - qualitative
2.1 For $y'=y-3$ , identify the equilibrium, sign regions, and long-run direction for initial values 1, 3, and 5.
Detailed solution
Practice - field fingerprint
2.2 Describe how the fields for $y'=\cos t$ , $y'=\cos y$ , and $y'=\cos(t+y)$ differ in repeated pattern.
Detailed solution
Integrated - derivatives
2.3 A solution satisfies $y'=y(1-y)$ and $y(0)=1/4$ . Predict monotonicity, limiting state, and the sign of $y''$ before and after $y=1/2$ .
Detailed solution
Challenge - nonuniqueness
2.4 Verify that both $y=0$ and $y=(t/2)^2$ for $t\ge0$ solve $y'=\sqrt{y}$ with $y(0)=0$ . What field-sketching rule must you suspend at this nonsmooth point?
Detailed solution
Practice - extension
2.5 For $y'=t-y$ , mark the zero-slope line and sign regions. Is the line $y=t$ itself a solution? For the solution through $(0,0)$ , compute $y'(0)$ and $y''(0)$ , then predict its initial motion for small positive time.
Detailed solution
Challenge - transfer
2.6 For $y'=t^2-y$ , derive $y''$ along a solution. A solution passes through $(1,1)$ , while another passes through $(-1,1)$ . Both have zero slope there. Determine which point is a local minimum and which is a local maximum, and explain why horizontal slope alone does not settle the question.
Detailed solution

Spaced retrieval 1

Without looking back, translate: "A quantity moves toward 10 at a rate equal to twice its signed distance from 10." Check the sign by testing a state above 10.

Reveal after committing

$y'=-2(y-10)=2(10-y)$ . At $y=12$ , the rate is $-4$ , so the state moves down toward 10.

I manipulated each field, made predictions first, and completed the spaced retrieval.

Next: Learn the structural fingerprints that select a solution method.

Choose a method from fingerprints

Chapter 3 - Recognition before execution

Solving is not only performing a method. It is deciding which method fits. Homework grouped by method hides that decision because every exercise under "Separable Equations" is separable. Real problems do not carry those labels.

This chapter builds a method-selection guide you will retrieve and refine throughout the book.

Normalize, then inspect

Before naming a method:

identify the unknown function and independent variable;
identify the highest derivative and therefore the order;
move or divide terms to expose a familiar form; and
check whether a simpler qualitative answer should come first.

The same equation can wear a disguise. For example,

$(1+t)y'=2y$

looks different from $y'=a(t)b(y)$ until division gives

$y'=\frac{2}{1+t}y.$

Now its separable fingerprint is visible. It is also linear; either method works.

The first-course method map

Autonomousy' = f(y)

First draw the phase line: a state axis with arrows showing motion. Equilibria and their response to nearby starting states may answer the important question without a formula.

Separabley' = a(t)b(y)

Move all state factors beside $dy$ and all time factors beside $dt$ .

First-order lineary' + p(t)y = q(t)

Use an integrating factor, a multiplier that creates one product derivative.

Second-order constant coefficientay'' + by' + cy = g(t)

Use characteristic roots, the rates of the equation's exponential solutions, for natural motion; add a response shaped like the input.

Linear systemx' = Ax

Read eigen-directions and eigenvalues, which encode basic motions such as growth, decay, and oscillation. Then connect them to a phase portrait: a picture of paths through state space.

No elementary fingerprinty' = f(t,y)

Use a field, numerical approximation, bounds, or a qualitative argument.

These categories overlap. $y'=ty$ is autonomous? No, because the rate contains $t$ . It is both separable and linear. Choose the shorter route and retain the other as a check.

Lab: classify before solving

Method trainerEight mixed equations; no chapter labels.

Contrast cases: train the boundary

Deep recognition grows when similar-looking examples require different moves.

Separable

$y'=t(1+y)$

$\dfrac{dy}{1+y}=t\,dt$

Why

one factor uses only $t$ , the other only $y$

Linear, not separable

$y'=t+y$

$y'-y=t$

Why

the sum $t+y$ cannot split into $a(t)b(y)$

Neither elementary form

$y'=t+y^2$

use a field or numerical method

Why

nonlinear in $y$ , and no product separates

Almost separable. Which equation fails the separable test?

Worked example: name two routes

Classify $ty'+2y=t^3$ , $t\gt 0$ method selection

Normalize. Divide by $t$ : $y'+(2/t)y=t^2$ .
Test separability. Rewriting gives $y'=t^2-(2/t)y$ , a sum whose terms mix $t$ and $y$ . It does not split into $a(t)b(y)$ .
Test linearity. It matches $y'+p(t)y=q(t)$ with $p(t)=2/t$ and $q(t)=t^2$ .
State the next move. Use the integrating factor $\mu=e^{\int 2/t\,dt}=t^2$ on $t\gt 0$ . Chapter 5 develops the method.

Fade the classification

Classify each equation and state only the first move, not the full solution.

$y'=(t^2+1)e^{-y}$
$y'-\frac{1}{t}y=t$
$y'=y^2-4$
$y''-5y'+6y=0$

Check the fingerprints

Separable; multiply by $e^y$ and pair $e^y dy$ with $(t^2+1)dt$ .
First-order linear; compute an integrating factor on an interval excluding zero.
Autonomous and separable; first find equilibria $y=\pm2$ and draw a phase line.
Homogeneous second-order constant coefficient; form $r^2-5r+6=0$ .

Method choice is a claim you can test

Before executing, point to the exact fingerprint. "It looks linear" is too vague. Say "after dividing by $t$ , the equation has $y'+p(t)y=q(t)$ , and $y$ appears only to the first power."

Error clinic: distributing $dy$ as if it were an algebra tile

Separation is justified through differential or integral notation, but only after the rate factors correctly. From $y'=t+y$ , writing $dy/y=t\,dt+dt$ does not separate anything: the right side still came from an invalid division of a sum.

Practice

For every problem, write a one-line fingerprint justification before any algebra.

Practice - classification
3.1 Classify: (a) $y'=t^2/y$ , (b) $y'+e^t y=1$ , (c) $y'=y\sin y$ , (d) $y''+9y=0$ .
Detailed solution
Practice - contrast
3.2 Explain why $y'=t y+t$ is separable, while $y'=t y+1$ is generally not. Identify a method that works for both.
Detailed solution
Integrated - multiple routes
3.3 Find every applicable early-course method for $y'=y/t$ , $t\gt 0$ . Which route uses less machinery?
Detailed solution
Challenge - strategy boundary
3.4 For $y'=y^2+t$ , explain precisely why the separable and first-order linear fingerprints fail. List three useful things you can still do without an elementary formula.
Detailed solution
Practice - extension
3.5 Normalize $(t-2)y'+3y=(t-2)^2$ . Identify the method and coefficient functions on intervals avoiding 2. Which interval containing $t=3$ is appropriate for the normalized equation? What condition would the original equation impose if a differentiable solution extended to $t=2$ ? Do not solve.
Detailed solution
Challenge - transfer
3.6 The equation $y'=y+ty^2$ is neither first-order linear nor separable in its given variables. Record the equilibrium lost by dividing by $y$ , then set $v=1/y$ on a nonzero branch. Derive the equation for $v$ and identify its method. Explain why the transformed equation does not make the original equation linear in $y$ .
Detailed solution

Retrieve the map

On blank paper, recreate the six-box method map from memory. For each box, write a representative equation that did not appear on this page. Then compare.

I rebuilt the method map from memory and classified the practice before solving anything.

Next: Turn the separable fingerprint into a reliable sequence of moves.

Separable equations

Chapter 4 - Put each variable on its own side

A first-order equation is separable when its rate factors into a time-only part and a state-only part:

$y'=a(t)b(y).$

The method reverses the chain rule. It rewrites the equation so one integral uses $y$ and the other uses $t$ .

The fingerprint and the move

See

$\dfrac{dy}{dt}=a(t)b(y)$

$\dfrac{1}{b(y)}\,dy=a(t)\,dt$

integrate both sides, then use the initial condition

Before dividing by $b(y)$ , solve $b(y)=0$ . Every root may give an equilibrium solution that division would erase.

Equilibrium safeguard

Write "lost-solution check: $b(y)=0$ " above the first line of every separation. This five-second habit prevents one of the most common errors in the course.

Worked example: full support

Solve $y'=ty$ , $y(0)=2$ fully worked

Recognize. The rate is $a(t)b(y)$ with $a(t)=t$ and $b(y)=y$ .
Protect equilibria. $b(y)=0$ gives $y=0$ , an equilibrium of the differential equation. It does not meet $y(0)=2$ , but it belongs to the general solution set.
Separate. On a region where $y\ne0$ , $\dfrac{1}{y}\,dy=t\,dt$ .
Integrate. $\ln|y|=t^2/2+C$ .
Expose the family. $|y|=e^Ce^{t^2/2}$ . Absorb sign and positive magnitude into one nonzero constant: $y=Ce^{t^2/2}$ . Including $C=0$ restores the equilibrium.
Use the condition. $2=C$ , so $\boxed{y=2e^{t^2/2}}$ .
Check. $y'=2te^{t^2/2}=ty$ , and $y(0)=2$ . The curve decreases for $t\lt 0$ , has zero slope at 0, and increases for $t\gt 0$ , exactly as $ty$ predicts for positive $y$ .

Why the absolute value disappears carefully

From $\ln|y|=F(t)+C$ , exponentiation gives $|y|=Ae^{F(t)}$ with $A\gt 0$ . A solution stays either positive or negative on the interval where the derivation holds. Write $y=Ce^{F(t)}$ , where $C\ne0$ can have either sign. Then check whether $C=0$ restores an equilibrium.

Erasing $|\cdot|$ without accounting for sign can delete negative solutions.

Fade the support backward

Solve $y'=(1+t)(1+y^2)$ , $y(0)=0$ .

Given

$\dfrac{dy}{1+y^2}=(1+t)dt$

$\arctan y=t+t^2/2+C$

Complete

Use $y(0)=0$ to find $C=\underline{\hspace{2em}}$ .

Apply tangent: $y=\underline{\hspace{8em}}$ .

Verify

Differentiate your function and name the interval containing 0 on which it remains finite.

Check the completion

$C=0$ , so $y=\tan(t+t^2/2)$ . The argument has minimum $-1/2$ , so it never reaches $-\pi/2$ . Both nearest poles instead occur where $t+t^2/2=\pi/2$ , at $t=-1\pm\sqrt{1+\pi}$ . The maximal interval containing 0 is $(-1-\sqrt{1+\pi},-1+\sqrt{1+\pi})$ . Substitution confirms $y'=(1+t)\sec^2(t+t^2/2)=(1+t)(1+y^2)$ .

Definite integrals keep conditions attached

Instead of adding constants and solving for them later, you can incorporate $y(t_0)=y_0$ directly:

$\int_{y_0}^{y(t)}\frac{1}{b(u)},du=\int_{t_0}^{t}a(s),ds.$

The dummy variables $u$ and $s$ prevent the same symbol from serving as both a limit and an integration variable. This form often reduces constant errors and naturally produces an implicit solution, an equation relating the variables without isolating the state.

Implicit can be finished

For $y'=t/(1+y^2)$ with $y(0)=1$ , definite integration gives

$\int_1^{y(t)}(1+u^2)du=\int_0^t s,ds,$

$y+\frac{y^3}{3}=\frac{t^2}{2}+\frac43.$

Solving the cubic explicitly would make the answer harder to read. The implicit equation already defines the solution locally and is easy to differentiate for verification.

Lost solution check. Separating $y'=y(1-y)$ requires division by $y(1-y)$ . Which solutions must be recorded first?

Error clinic

Incorrect:

y'=t+y \Rightarrow dy/y=t\,dt+dt

Division by $y$ gives $y'/y=t/y+1$ , not $t+1$ . A sum does not separate unless the entire right side factors into a time-only part times a state-only part.

Incomplete:

\int y^{-1}dy=\ln y

The real antiderivative is $\ln|y|$ on intervals avoiding zero. If an initial condition forces $y\gt 0$ , say so; otherwise preserve both signs.

Practice

Practice - core move
4.1 Solve $y'=3t^2/y$ , $y(0)=2$ . State the branch selected by the initial condition and verify it.
Detailed solution
Practice - equilibrium safeguard
4.2 Find all equilibrium and non-equilibrium solutions of $y'=y(y-2)$ . Then solve the initial-value problem $y(0)=1$ .
Detailed solution
Integrated - modeling
4.3 A culture grows at rate $P'=0.6P(1-P/500)$ , with $P(0)=50$ . Separate and solve. Use the equation, not only the formula, to predict the long-run population.
Detailed solution
Challenge - interval
4.4 Solve $y'=1+y^2$ , $y(0)=1$ . Find the maximal open interval containing zero on which the solution exists.
Detailed solution
Practice - extension
4.5 Solve $y'=2ty^2$ , $y(0)=-2$ . Include the lost-solution check, state the maximal interval containing zero, and explain why the solution stays negative even though its derivative changes sign.
Detailed solution
Challenge - transfer
4.6 Solve $y'=y^{3/2}$ , $y(0)=4$ , for real nonnegative $y$ . State the maximal interval containing zero and verify the branch restriction before squaring. Explain why the resulting rational expression is not a continuation of this solution past its pole.
Detailed solution

Spaced retrieval 2

Without notes:

rebuild the three-pass verification from Chapter 1;
state the horizontal-row fingerprint from Chapter 2; and
write the equilibrium safeguard for separation.

I solved the faded example on paper and retrieved the three earlier ideas.

Next: Use an integrating factor to turn a linear equation into one product derivative.

First-order linear equations

Chapter 5 - Manufacture a product derivative

A first-order linear equation has the normalized form

$y'+p(t)y=q(t).$

The obstacle is the extra $p(t)y$ term. An integrating factor $\mu(t)$ makes the left side become one product derivative:

$\mu y'+\mu p y=(\mu y)'.$

Derive the method once

The product rule says

$(\mu y)'=\mu y'+\mu' y.$

To match $\mu y'+\mu p y$ , require $\mu'=\mu p$ , or $\mu'/\mu=p$ . Therefore

$\boxed{\mu(t)=e^{\int p(t),dt}}.$

This formula is not a magic insertion. It solves the small design problem "which multiplier makes these two terms a product derivative?"

The multiplier is chosen to create structure, not merely to change the equation.

The reliable sequence

Normalize the coefficient of $y'$ to 1.
Identify $p(t)$ and $q(t)$ .
Compute $\mu=e^{\int p(t)dt}$ .
Multiply every term by $\mu$ .
Replace the left side with $(\mu y)'$ .
Integrate and solve for $y$ .
Apply the initial condition and verify.

Worked example: full support

Solve $y'+2y=6$ , $y(0)=5$ fully worked

Read the model. $y'=6-2y=-2(y-3)$ . Before solving, predict movement toward the equilibrium $y=3$ . Starting at 5, the curve should decrease and flatten.
Identify. $p(t)=2$ , $q(t)=6$ .
Build the factor. $\mu=e^{\int 2dt}=e^{2t}$ .
Multiply all terms. $e^{2t}y'+2e^{2t}y=6e^{2t}$ .
Recognize the product. $(e^{2t}y)'=6e^{2t}$ .
Integrate. $e^{2t}y=3e^{2t}+C$ , hence $y=3+Ce^{-2t}$ .
Use and check. $5=3+C$ , so $y=3+2e^{-2t}$ . Its graph decreases toward 3, matching the prediction. Substitution gives $y'+2y=-4e^{-2t}+6+4e^{-2t}=6$ .

Backward-faded example

Solve $ty'-y=t^3$ , $t\gt 0$ , with $y(1)=2$ .

Given

Normalize: $y'-\frac1t y=t^2$ .

Thus $p(t)=-1/t$ .

Complete

$\mu=e^{\int -1/t\,dt}=\underline{\hspace{4em}}$ .

Then $(\underline{\hspace{3em}}\,y)'=\underline{\hspace{4em}}$ .

You finish

Integrate, use $y(1)=2$ , and verify in the normalized equation.

Check the missing work

On $t\gt 0$ , $\mu=e^{-\ln t}=1/t$ . Multiplication gives $(y/t)'=t$ . Thus $y/t=t^2/2+C$ , so $y=t^3/2+Ct$ . The condition gives $C=3/2$ , hence $y=(t^3+3t)/2$ .

When linear and separable overlap

The homogeneous linear equation $y'+p(t)y=0$ , with no external input on the right, is also separable:

$\frac{y'}{y}=-p(t).$

For $y'=ty$ , separation is shorter. For $y'+p(t)y=q(t)$ with nonzero $q$ , the integrating-factor route usually survives while separation fails.

Normalize first. For $2ty'+6y=t^2$ on $t\gt 0$ , what is $p(t)$ ?

Modeling example: a mixing tank

A 100 L tank receives brine at 2 L/min with concentration 0.1 kg/L. The same volume leaves each minute. If $S(t)$ is kilograms of salt, then

$S'=\underbrace{(2)(0.1)}$

In linear form, $S'+0.02S=0.2$ . Before solving, the zero-rate amount is 10 kg, so every solution should move toward 10. The integrating factor gives $S=10+Ce^{-0.02t}$ . The constant records the initial displacement from equilibrium.

Error clinic

Error clinic: reading $p(t)$ before normalization

In $t y'+2y=1$ , the coefficient used in the integrating factor is $2/t$ , not 2. Divide by the coefficient of $y'$ first and state the interval, such as $t\gt 0$ .

Error clinic: multiplying only the left side

An integrating factor preserves equality only when it multiplies every term. Write the multiplied equation in full before compressing the product derivative.

Practice

Practice - constant coefficient
5.1 Solve $y'-3y=6$ , $y(0)=1$ . Predict the long-run behavior first and reconcile it with the result.
Detailed solution
Practice - variable coefficient
5.2 Solve $y'+(2/t)y=t$ for $t\gt 0$ , with $y(1)=0$ .
Detailed solution
Integrated - model
5.3 A room at 20 °C contains an object initially at 80 °C. Newton cooling gives $T'=-0.25(T-20)$ . Solve, then find when the object reaches 30 °C.
Detailed solution
Challenge - reverse engineering
5.4 Find a first-order linear differential equation whose general solution is $y=t^2+Ce^{-t}$ . Explain how your equation separates the forced response from the transient.
Detailed solution
Practice - extension
5.5 Solve $y'+2y=e^{-t}$ , $y(0)=0$ , using an integrating factor. For $t\ge0$ , determine when the solution reaches its maximum and find that maximum. Check both the equation and the initial condition.
Detailed solution
Challenge - transfer
5.6 Solve the variable-volume tank model $S'+S/(100+t)=6$ , $S(0)=0$ , for $t\ge0$ , with $S$ in grams and $100+t$ in liters. Find the concentration $S/(100+t)$ and its limit. Explain why salt amount has no finite limit even though concentration does.
Detailed solution

Retrieve and discriminate

On blank paper, solve only far enough to expose the key transformation:

$y'=t(1-y)$ by separation;
$y'+ty=t$ by an integrating factor; and
show that these are the same equation, so both methods apply; then explain why replacing the right side of the second equation by $t^2$ generally removes separability but preserves linearity.

I derived the integrating factor from the product rule and completed the faded solution.

Next: Use phase lines and feedback to understand autonomous equations.

Autonomous equations and stability

Chapter 6 - Read feedback from a phase line

An autonomous equation has no explicit time dependence:

$y'=f(y).$

The same state always receives the same rate. This makes long-run behavior visible on a one-dimensional phase line, often before any solution formula appears.

The phase-line procedure

Solve $f(y)=0$ for equilibrium states.
Use those states to divide the line into intervals.
Test the sign of $f(y)$ on each interval.
Draw an arrow up for $f(y)\gt 0$ and down for $f(y)\lt 0$ .
Classify each equilibrium by the arrows around it.

An equilibrium is:

asymptotically stable when nearby arrows point toward it (called stable in the phase-line examples below);
unstable when nearby arrows point away; and
semistable when one side points in and the other points out.

A phase line compresses every solution's direction into one state axis.

Lab: logistic feedback

Logistic feedback explorerChange the environment and the initial state.

Growth rate r 0.8 Capacity K 6.0 Initial y(0) 1.5

solutioncarrying capacity

Questions to test in the lab:

Which parameter changes the destination?
Which parameter changes how quickly the solution travels?
Does a positive initial state above $K$ blow up, cross $K$ , or approach $K$ ?

Worked example: stability before formula

Analyze $y'=y(2-y)(y-5)$ qualitative first

Equilibria. The zeros are $y=0,2,5$ .
Sign below 0. Test $y=-1$ : $(-)(+)(-)$ is positive, so arrows point up.
Sign from 0 to 2. Test $y=1$ : $(+)(+)(-)$ is negative, so arrows point down.
Sign from 2 to 5. Test $y=3$ : $(+)(-)(-)$ is positive, so arrows point up.
Sign above 5. All factors have signs $(+)(-)(+)$ , so the rate is negative.
Classify. Arrows point toward 0, away from 2, and toward 5. Thus 0 and 5 are stable; 2 is unstable.

Fade the phase line

For $y'=(y+1)(y-2)^2$ :

equilibria: $y=\underline{\hspace{4em}}$ ;
sign on $( -\infty,-1)$ : $\underline{\hspace{2em}}$ ;
sign on $(-1,2)$ : $\underline{\hspace{2em}}$ ;
sign on $(2,\infty)$ : $\underline{\hspace{2em}}$ ;
stability at $-1$ : $\underline{\hspace{5em}}$ ; at 2: $\underline{\hspace{5em}}$ .

Check the line

The equilibria are $-1$ and 2. Since the squared factor never changes sign, the rate is negative below $-1$ and positive on both intervals above $-1$ . Thus $-1$ is unstable, while 2 is semistable: trajectories below 2 move upward toward it, while trajectories above 2 move upward away from it.

Feedback language

Near an equilibrium $y_*$ , examine whether a small displacement triggers a correcting or amplifying rate.

Negative feedback: a positive displacement produces a negative rate, and a negative displacement produces a positive rate. The equilibrium restores itself.
Positive feedback: a displacement produces a rate in the same direction. The equilibrium repels.

When $f$ is differentiable and $f'(y_*)\ne0$ :

$f'(y_*)\lt 0$ suggests local stability;
$f'(y_*)\gt 0$ suggests local instability.

The sign chart remains safer when $f'(y_*)=0$ .

Stability from feedback. Near $y=3$ , a model is approximately $y'=-0.4(y-3)$ . What happens after a small upward displacement?

Parameters can change the portrait

Consider $y'=y(r-y)$ . When $r\gt 0$ , $y=0$ is unstable and $y=r$ is stable. When $r\lt 0$ , their order swaps: $y=r$ is unstable and $y=0$ is stable. At $r=0$ , the equilibria meet.

This qualitative change as a parameter passes a threshold is a bifurcation. It matters because parameter changes can alter not only speed, but the number and stability of long-run states.

Error clinic

Error clinic: putting time on the phase line

A phase line's coordinate is the state $y$ . Arrows report how $y$ changes as time advances. A solution graph uses horizontal time and vertical state; the two pictures carry different axes.

Practice

Practice - phase line
6.1 Draw and classify the phase line for $y'=y(y-1)(y-3)$ .
Detailed solution
Practice - solution behavior
6.2 For the same equation, describe the long-run behavior for $y(0)=-1, 1/2, 2, 4$ . Do not solve.
Detailed solution
Integrated - logistic interpretation
6.3 In $P'=rP(1-P/K)$ , with $r,K\gt 0$ , show that the growth rate $P'$ is largest at $P=K/2$ . Distinguish "largest growth rate" from "largest population."
Detailed solution
Challenge - threshold
6.4 Analyze $y'=y^2-a$ for $a\gt 0$ , $a=0$ , and $a\lt 0$ . Track how the equilibria and their stability change.
Detailed solution
Practice - extension
6.5 Analyze $y'=-(y-2)^3$ . Classify the equilibrium using a sign chart, then compute $f'(2)$ . What does this example show about concluding instability from a zero linearization derivative? Predict the limits for $y(0)=1$ and $y(0)=3$ .
Detailed solution
Challenge - transfer
6.6 A harvested population follows $P'=P(1-P/8)-H$ , where $H\ge0$ is a constant removal rate. Find and classify equilibria for $0\le H\lt 2$ , $H=2$ , and $H\gt 2$ . For $0\lt H\lt 2$ , explain the role of the smaller equilibrium. Treat $P=0$ as extinction: stop the physical model there instead of allowing negative population.
Detailed solution

Retrieve and connect

Without looking, answer:

Why must separation record equilibria before division?
How does an integrating factor use the product rule?
What do inward, outward, and one-sided arrows mean at an equilibrium?

I drew each phase line before doing algebra and retrieved the two earlier methods.

Next: Turn second-order equations into the language of natural modes and oscillation.

Second-order equations and oscillation

Chapter 7 - Position needs velocity

A second-order equation tracks curvature or acceleration. To select one motion, you usually need two initial facts, such as position and velocity:

$y''+ay'+by=0,\qquad y(0)=y_0,\quad y'(0)=v_0.$

The solution is built from natural modes: motions the unforced system can sustain on its own.

From derivatives to a polynomial

Exponentials reproduce themselves under differentiation. Try $y=e^{rt}$ :

$r^2e^{rt}+ar e^{rt}+be^{rt}=0.$

Since $e^{rt}\ne0$ , the possible rates $r$ solve the characteristic equation

$r^2+ar+b=0.$

The roots predict the motion's geometry.

Two real roots $r_1\ne r_2$ y=C_1e^{r_1t}+C_2e^{r_2t}

The mode with the larger root dominates late if its coefficient is nonzero. If both roots are negative, this is the slower-decaying mode.

Repeated real root $r$ y=(C_1+C_2t)e^{rt}

The factor $t$ creates a second independent motion.

Complex roots $\alpha\pm i\beta$ y=e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t)

$\alpha$ controls the envelope, the curve bounding the oscillation's size; $\beta$ controls angular frequency, measured in radians per unit time.

Root signsRe(r) < 0 → decay

Positive real part gives a growing mode. A nonzero pure imaginary pair gives undamped oscillation; a zero root gives a constant mode, and a repeated zero root also permits linear drift.

Lab: watch damping reshape motion

Damped oscillator

x''+cx'+x=0

Damping c 0.35 Initial position 1.5 Initial velocity 0.0

position over timeposition-velocity path

Move $c$ slowly through 2. For $0\lt c\lt 2$ , roots are complex and nonzero trajectories spiral inward; at $c=0$ , they form closed orbits without decay. At 2, the root repeats and the system returns without sustained oscillation. Above 2, two negative real modes give a non-oscillatory return; the slower mode decays more slowly than at critical damping unless its coefficient is zero. Critical damping is the boundary between oscillatory and non-oscillatory return.

Worked example: read roots, then constants

Solve $y''+4y'+13y=0$ , $y(0)=2$ , $y'(0)=-1$ fully worked

Characteristic equation. $r^2+4r+13=0$ .
Roots. $r=(-4\pm\sqrt{16-52})/2=-2\pm3i$ .
Predict. The motion oscillates at angular frequency 3 under an envelope $e^{-2t}$ . It decays rapidly.
Write the real family. $y=e^{-2t}(C_1\cos3t+C_2\sin3t)$ .
Use position. $y(0)=C_1=2$ .
Use velocity. Differentiating and evaluating at zero gives $y'(0)=-2C_1+3C_2=-1$ . With $C_1=2$ , $C_2=1$ .
Result and check. $y=e^{-2t}(2\cos3t+\sin3t)$ . The roots guarantee the equation; direct evaluation checks both initial conditions.

Fade the root cases

Match each equation to a root picture and solution form.

$y''+5y'+6y=0$

Roots: $\underline{\hspace{5em}}$

Form: $\underline{\hspace{8em}}$

$y''+4y'+4y=0$

Root: $\underline{\hspace{5em}}$

Form: $\underline{\hspace{8em}}$

$y''+9y=0$

Roots: $\underline{\hspace{5em}}$

Interpret the motion.

Check the forms

Roots $-2,-3$ ; $y=C_1e^{-2t}+C_2e^{-3t}$ .
Repeated root $-2$ ; $y=(C_1+C_2t)e^{-2t}$ .
Roots $\pm3i$ ; $y=C_1\cos3t+C_2\sin3t$ , an undamped oscillation.

Mechanical language

The mass-spring-damper model

$mx''+cx'+kx=0$

balances inertia $mx''$ , damping $cx'$ , and restoring force $kx$ . Dividing by $m$ shows that the key comparisons use $c/m$ and $k/m$ . The discriminant $c^2-4mk$ separates overdamped, critically damped, and underdamped motion.

Root reading. A characteristic root is $-0.2+5i$ . What does each part control?

Error clinic

Error clinic: one constant for a second-order equation

A second-order linear homogeneous equation generally needs two independent modes and two constants. A repeated root does not reduce that dimension; it changes the second mode to $te^{rt}$ .

Error clinic: using complex exponentials but dropping real solutions

Complex roots are a calculation device. For real initial data, report the equivalent real sine-cosine form unless the context explicitly prefers complex notation.

Practice

Practice - distinct roots
7.1 Solve $y''-y'-6y=0$ , $y(0)=1$ , $y'(0)=0$ . Predict which mode dominates for large positive time.
Detailed solution
Practice - repeated root
7.2 Solve $y''+6y'+9y=0$ , $y(0)=0$ , $y'(0)=2$ .
Detailed solution
Integrated - oscillator
7.3 For $x''+2\zeta x'+x=0$ , classify motion for $\zeta=0, 1/2, 1, 2$ . Connect each case to a path in position-velocity space.
Detailed solution
Challenge - design
7.4 Choose $a$ and $b$ so $y''+ay'+by=0$ has oscillations with envelope $e^{-t}$ and period $\pi$ . Explain from the desired roots.
Detailed solution
Practice - extension
7.5 Solve $x''+4x=0$ , $x(0)=1$ , $x'(0)=2$ . Rewrite the solution as a single shifted cosine, identify its amplitude and period, and verify that $E=(x')^2/2+2x^2$ stays constant.
Detailed solution
Challenge - transfer
7.6 Solve $y''+2y'=0$ , $y(0)=1$ , $y'(0)=-4$ . Identify both characteristic modes and the limit as $t\to\infty$ . Explain why the zero root produces neither a decaying mode nor a sustained oscillation.
Detailed solution

Retrieve across chapters

Explain how these are the same idea in different dimensions:

a stable equilibrium on a phase line;
a negative real characteristic root; and
an inward-moving position-velocity path.

I predicted motion from roots before solving for constants and connected the two plots in the lab.

Next: Add an external input and learn why matching frequencies can amplify motion.

Forcing and resonance

Chapter 8 - Separate natural motion from response

A nonhomogeneous equation adds an input:

$ay''+by'+cy=g(t).$

Every solution splits into

$\boxed{y=y_h+y_p},$

where $y_h$ is homogeneous natural motion and $y_p$ is one particular response to the forcing. Natural motion is a decaying transient only when all homogeneous modes decay; undamped oscillations persist, and unstable modes can grow.

Match the input's shape

For constant coefficients, undetermined coefficients proposes a particular solution with the same family of shapes as $g(t)$ .

Polynomialg(t)=t²+1

Try a general polynomial of the same degree.

Exponentialg(t)=e^{at}

Try $Ae^{at}$ .

Sinusoidg(t)=cos(ωt)

Try $A\cos(ωt)+B\sin(ωt)$ .

Productsg(t)=te^{at}

Multiply the corresponding trial families.

If the trial already belongs to $y_h$ , multiply it by enough powers of $t$ to make it independent. This is the algebraic signature of resonance.

Worked example: ordinary forcing

Solve $y''+y=3\cos2t$ shape matching

Natural motion. $r^2+1=0$ gives $y_h=C_1\cos t+C_2\sin t$ .
Choose a trial. Since frequency 2 does not appear in $y_h$ , try $y_p=A\cos2t+B\sin2t$ .
Substitute. $y_p''+y_p=(-4A+A)\cos2t+(-4B+B)\sin2t$ .
Match coefficients. $-3A=3$ and $-3B=0$ , so $A=-1$ , $B=0$ .
Combine. $y=C_1\cos t+C_2\sin t-\cos2t$ .
Interpret. The output contains natural frequency 1 and forced frequency 2. Initial conditions would set the natural-motion amplitudes.

Lab: sweep through resonance

Forced oscillator

x''+cx'+x=\cos(\omega t)

Forcing frequency ω 0.70 Damping c 0.18

time responsesteady-state amplitude

Reduce damping, then move $\omega$ near 1. The response curve rises sharply because the input frequency approaches the natural frequency. Damping keeps the steady-state amplitude finite and shifts the exact peak slightly below 1.

Resonant forcing

For

$y''+y=\cos t,$

the ordinary trial $A\cos t+B\sin t$ duplicates $y_h$ . Multiply by $t$ :

$y_p=t(A\cos t+B\sin t).$

Substitution gives one valid choice $y_p=\tfrac12 t\sin t$ . Its envelope grows linearly because an undamped input adds energy in phase with the natural motion.

Trial collision. For $y''-2y'+y=e^t$ , which particular trial is appropriate?

The root $r=1$ repeats twice in the homogeneous equation, so multiply the exponential trial by $t^2$ .

Fade the response

Find a particular solution of $y''+3y'+2y=4e^{-t}$ .

Characteristic roots: $\underline{\hspace{5em}}$ .
Collision? The input $e^{-t}$ [is / is not] a natural mode.
Correct trial: $y_p=\underline{\hspace{6em}}$ .
Substitute and find the coefficient.

Check the response

The roots are $-1,-2$ , so $e^{-t}$ collides once. Try $y_p=Ate^{-t}$ . Substitution yields $A e^{-t}=4e^{-t}$ , so $A=4$ . One particular solution is $4te^{-t}$ .

Transient versus steady state

In a damped stable system, $y_h$ decays. What remains is the steady-state response $y_p$ . This split explains why two experiments with different initial conditions can eventually show the same driven behavior.

Error clinic: applying initial conditions to $y_p$ alone

The particular solution only supplies one response to the input. Initial conditions select constants after $y_h+y_p$ has been assembled.

Practice

Practice - polynomial input
8.1 Find the general solution of $y''+y=t$ .
Detailed solution
Practice - sinusoidal input
8.2 Solve $y''+4y=\cos t$ , $y(0)=0$ , $y'(0)=0$ .
Detailed solution
Integrated - collision
8.3 For each input to $y''+4y= g(t)$ , state the trial without solving: (a) $e^{-t}$ , (b) $\cos2t$ , (c) $t^2$ .
Detailed solution
Challenge - beating
8.4 Use $\cos at-\cos bt=-2\sin((a+b)t/2)\sin((a-b)t/2)$ to explain the slow amplitude modulation when an undamped oscillator is forced near, but not at, its natural frequency.
Detailed solution
Practice - extension
8.5 Solve $y''-2y'+y=e^t$ , $y(0)=0$ , $y'(0)=0$ . Explain the required power of $t$ in the particular trial, and verify the complete solution using the substitution $y=e^t u$ .
Detailed solution
Challenge - transfer
8.6 Solve $y''+2y'+2y=\cos t$ , $y(0)=0$ , $y'(0)=0$ . Find the steady-state amplitude, identify the transient, and explain why two solutions with different initial conditions approach the same periodic response.
Detailed solution

Retrieve the design logic

Explain without formulas:

why a particular solution should resemble the input;
why a collision with natural motion requires a factor of $t$ ; and
why damping changes a resonance peak.

I swept the resonance lab, predicted each trial shape, and explained every collision.

Next: Read coupled states as vector fields, eigen-directions, and phase portraits.

Systems and phase portraits

Chapter 9 - Several states move together

A system tracks a state vector, a list of state components, rather than one number:

$\mathbf{x}'=A\mathbf{x},\qquad \mathbf{x}=\begin{bmatrix}x\y\end{bmatrix}.$

At every point in the phase plane, whose axes are the two state components, the system assigns a velocity vector. A solution follows a trajectory, a path tangent to that field.

Lab: change the eigenvalue pattern

Linear phase-plane explorerClick to launch another trajectory.

System type

velocity fieldtrajectoriesequilibrium

Ask three questions for each preset:

Do trajectories approach or leave the origin?
Do they rotate?
Are any straight-line directions preserved?

Eigenvectors are invariant directions

If $A\mathbf{v}=\lambda\mathbf{v}$ , try a solution $\mathbf{x}=e^{\lambda t}\mathbf{v}$ . Then

$\mathbf{x}'=\lambda e^{\lambda t}\mathbf{v} =e^{\lambda t}A\mathbf{v} =A\mathbf{x}.$

Starting on a real eigenvector line keeps the trajectory on that line. Its real eigenvalue says whether motion grows, decays, or stays constant. A complex conjugate pair describes rotating motion in the real plane, not two real eigenvector lines.

Both real parts negativestable node or spiral sink

Nearby trajectories approach the equilibrium.

Both real parts positiveunstable node or spiral source

Nearby trajectories leave the equilibrium.

Opposite signssaddle

One direction contracts while another expands.

Pure imaginary paircenter in the ideal linear case

Trajectories orbit without decay or growth.

Worked example: an unstable node

Analyze $x'=3x+y,\ y'=x+3y$ eigen-directions

Matrix. $A=\begin{bmatrix}3&1\\1&3\end{bmatrix}$ .
Eigenvalues. $\det(A-\lambda I)=(3-\lambda)^2-1=0$ , so $\lambda=4,2$ .
Eigenvectors. For $\lambda=4$ , one eigenvector is $[1,1]^T$ . For $\lambda=2$ , one is $[1,-1]^T$ .
Build the family. $\mathbf{x}=C_1e^{4t}[1,1]^T+C_2e^{2t}[1,-1]^T$ .
Classify. Both eigenvalues are positive, so the origin is an unstable node, not a saddle. The $\lambda=4$ direction dominates for most initial states as $t\to\infty$ .
Sketch. Draw both eigenvector lines first. Other trajectories leave the origin and bend toward the faster $[1,1]^T$ direction.

Classification comes from eigenvalue signs, not from the visual appearance of the matrix. A saddle would require eigenvalues with opposite signs.

Backward-faded system

For $x'=x,\ y'=-2y$ :

matrix: $A=\underline{\hspace{7em}}$ ;
eigenpairs: $(\lambda_1,\mathbf v_1)=\underline{\hspace{7em}}$ , $(\lambda_2,\mathbf v_2)=\underline{\hspace{7em}}$ ;
general solution: $x=\underline{\hspace{4em}}$ , $y=\underline{\hspace{4em}}$ ;
classification: $\underline{\hspace{5em}}$ .

Check the portrait

$A=\operatorname{diag}(1,-2)$ . Eigenpairs are $(1,[1,0]^T)$ and $(-2,[0,1]^T)$ . Thus $x=C_1e^t$ , $y=C_2e^{-2t}$ . One direction grows and one decays, so the origin is a saddle.

Second-order equations are systems in disguise

Let $x_1=y$ and $x_2=y'$ . Then

$y''+ay'+by=0$

becomes

$\begin{bmatrix}x_1'\x_2'\end{bmatrix} = \begin{bmatrix}0&1\-b&-a\end{bmatrix} \begin{bmatrix}x_1\x_2\end{bmatrix}.$

The characteristic polynomial of this matrix is the same $\lambda^2+a\lambda+b$ from Chapter 7. The time trace and phase portrait are two views of one solution.

Classification. A 2×2 system has eigenvalues $-1$ and $3$ . What is the origin?

Error clinic

Error clinic: drawing component graphs in the phase plane

The phase plane plots $y(t)$ against $x(t)$ ; time is implicit along the trajectory. A time-series plot uses $t$ on the horizontal axis. Label axes before interpreting a curve.

Error clinic: classifying from trace alone

A negative trace does not guarantee stability when the determinant is negative. Opposite-sign eigenvalues always produce a saddle. Compute enough information to determine both eigenvalue signs or real parts.

Practice

Practice - diagonal system
9.1 Solve and sketch $x'=-x,\ y'=-3y$ . Which eigen-direction dominates late?
Detailed solution
Practice - coupled system
9.2 Find eigenvalues, eigenvectors, and the general solution for $\mathbf{x}'=\begin{bmatrix}2&1\\1&2\end{bmatrix}\mathbf{x}$ .
Detailed solution
Integrated - oscillator bridge
9.3 Convert $y''+2y'+5y=0$ into a first-order system. Use its eigenvalues to predict the phase portrait.
Detailed solution
Challenge - trace and determinant
9.4 For a real 2×2 matrix, eigenvalues satisfy $\lambda^2-\tau\lambda+\Delta=0$ , where $\tau$ is trace and $\Delta$ determinant. Build a qualitative classification map in the $(\tau,\Delta)$ -plane.
Detailed solution
Practice - extension
9.5 Solve $x'=-x+y,\ y'=-y$ , with $x(0)=0,\ y(0)=1$ , by finding $y$ first. Compute the matrix's eigenvalues and real eigenvector directions. Explain why one eigenvector mode alone cannot satisfy these initial conditions, and classify the origin.
Detailed solution
Challenge - transfer
9.6 For $x'=-4y,\ y'=x$ , with $x(0)=2,\ y(0)=0$ , find the solution and prove that $x^2+4y^2$ is conserved. Determine the orbit, rotation direction, and period. Is the origin attracting?
Detailed solution

Retrieve the representation bridge

Choose one oscillator setting from Chapter 7. Describe the same motion in four languages: roots, time trace, phase portrait, and physical behavior.

I clicked new initial states, marked eigen-directions first, and translated between time and phase views.

Next: Approximate a solution step by step and learn to see numerical error.

Numerical solutions and error

Chapter 10 - Follow the local rule in finite steps

Many useful differential equations have no solution expressible by a finite combination of familiar elementary functions. Numerical methods turn the rate rule into a sequence of approximations.

Euler's method uses the current slope for one short step:

$\boxed{y_{n+1}=y_n+h f(t_n,y_n)},\qquad t_{n+1}=t_n+h.$

It is the tangent-line approximation repeated.

Lab: shrink the step

Euler step-size explorerCompare the polygonal approximation with the exact curve.

Equation Step h 0.60

exact solutionEuler approximation

For each equation, compare $h=1$ , $0.5$ , $0.25$ , and $0.1$ . Smaller steps usually reduce error, but they require more work. "Numerical" does not mean "unreliable"; it means error must be estimated and controlled.

Worked example: show every step

Approximate $y'=t-y$ , $y(0)=1$ , with $h=0.5$ three Euler steps

Start. $(t_0,y_0)=(0,1)$ . The slope is $f(0,1)=-1$ .
Step to $t=0.5$ . $y_1=1+0.5(-1)=0.5$ .
Recompute the slope. At $(0.5,0.5)$ , $f=0.5-0.5=0$ . Do not reuse the old slope.
Step to $t=1$ . $y_2=0.5+0.5(0)=0.5$ .
Recompute again. At $(1,0.5)$ , $f=0.5$ . Thus $y_3=0.5+0.5(0.5)=0.75$ at $t=1.5$ .
Behavior check. The state first falls because $t-y\lt 0$ , pauses near the line $y=t$ , then rises after $t-y\gt 0$ . The discrete values reproduce that qualitative change.

A table is part of the reasoning

$n$	$t_n$	$y_n$	slope $f(t_n,y_n)$	next $y$
0	0.0	1.000	-1.000	0.500
1	0.5	0.500	0.000	0.500
2	1.0	0.500	0.500	0.750

The table prevents a common error: evaluating the new slope at a mixture of old and new coordinates.

Local and global error

Euler replaces a curved segment by a tangent step. If the exact solution has bounded second derivative, one step's local truncation error (the unscaled one-step defect from exact starting data) is $O(h^2)$ . Some texts divide this defect by $h$ and call the result local truncation error instead. If $f$ is also Lipschitz in the state near the solution, accumulation across a fixed time interval gives global error $O(h)$ . These are upper-order bounds; special solutions can have zero error.

Practical consequence: halving $h$ should roughly halve Euler's global error once the steps are small enough for this small-step pattern to appear.

Error estimate. Euler with $h=0.2$ has error about 0.08 at a fixed final time. What error would first-order behavior predict for $h=0.1$ ?

Better slopes: midpoint and Runge-Kutta

Euler uses the slope at the start of a step. A midpoint method samples a slope near the middle. Classical fourth-order Runge-Kutta blends four slope samples. The central design idea is the same: spend more rate evaluations to approximate the curve across each step more accurately.

You do not need to memorize this fourth-order method, often called RK4, here. You should be able to ask:

where does the method sample slopes?
what order of global error does it claim?
does halving the step produce the expected convergence?
does the numerical curve respect known equilibria, signs, and bounds?

Numerical stability is not physical stability

For $y'=-10y$ , the true solution decays. Euler gives

$y_{n+1}=(1-10h)y_n.$

If $h=0.3$ , the multiplier is $-2$ : the approximation alternates and grows even though the true system decays. The step is too large for Euler's stability region.

Error clinic: trusting a smooth-looking plot

A plotting tool can connect inaccurate points smoothly. Test convergence by rerunning with a smaller step, and compare the result with qualitative facts from the differential equation.

Fade the computation

Use Euler with $h=0.25$ for $y'=y(1-y)$ , $y(0)=0.2$ .

$y_1=0.2+0.25[0.2(0.8)]=\underline{\hspace{4em}}$ .
Evaluate the next slope at $(t_1,y_1)=(0.25,\underline{\hspace{3em}})$ .
Compute $y_2$ and confirm that the approximation stays between 0 and 1.

Check two steps

$y_1=0.24$ . Then the slope is $0.24(0.76)=0.1824$ , so $y_2=0.24+0.25(0.1824)=0.2856$ . Both points respect the phase-line prediction of growth toward 1.

Practice

Practice - Euler table
10.1 Use Euler with $h=0.2$ to approximate $y(0.6)$ for $y'=y-t$ , $y(0)=1$ . Show a slope table.
Detailed solution
Practice - convergence
10.2 For $y'=y$ , $y(0)=1$ , derive Euler's approximation $y_n=(1+h)^n$ . Compare $h=1,1/2,1/4$ at $t=1$ with $e$ .
Detailed solution
Integrated - qualitative check
10.3 Design a test that would catch a numerical solution of $P'=P(1-P/10)$ , $P(0)=2$ , crossing above 10.
Detailed solution
Challenge - step stability
10.4 For $y'=-ay$ with $a\gt 0$ , find all $h\gt 0$ for which Euler approximations decay in magnitude. Which subset decays without alternating sign?
Detailed solution
Practice - extension
10.5 The explicit midpoint method uses $k_1=f(t_n,y_n)$ , $k_2=f(t_n+h/2,y_n+hk_1/2)$ , and $y_{n+1}=y_n+hk_2$ . Apply two steps with $h=0.5$ to $y'=y-t,\ y(0)=0$ . Compare $y(1)$ with two Euler steps and the exact value from $y=t+1-e^t$ .
Detailed solution
Challenge - transfer
10.6 For $z'=rz(1-z)$ , $r\gt 0$ , Euler gives $z_{n+1}=z_n+hrz_n(1-z_n)$ . Show that $0\lt z_n\lt 1$ implies $0\lt z_{n+1}\lt 1$ whenever $0\lt hr\le1$ . Test $r=1,\ h=2,\ z_0=0.75$ to demonstrate overshoot, and explain why a bound-preserving step is not automatically accurate.
Detailed solution

Retrieve the whole loop

Choose one equation from any earlier chapter and state:

its structural fingerprint;
a qualitative prediction;
an analytic method, if available;
a numerical fallback; and
two independent checks.

I compared at least four step sizes and checked the numerical results against qualitative behavior.

Next: Enter the mixed practice studio, where the method is never named for you.

Mixed practice studio

Chapter 11 - The method is part of the problem

This studio removes chapter labels. Your first written line for every problem must name a structural feature, a qualitative prediction, or both. Only then should you calculate.

How to use the studio

For each problem:

cover the solution link;
write a method and confidence from 0 to 100%;
solve or analyze;
verify with a different representation; and
if you miss, record the first wrong decision, not merely the final wrong answer.

Error log format

"I treated a sum as a product, so I chose separation" is useful. "Careless mistake" is not. Name the cue you missed and the cue you will inspect next time.

Round A: recognize and predict

Do not solve these. Name the most useful first method and one behavior or checking strategy.

Mixed - recognition
M1 $y'=ty+\sin t$
Detailed solution
Mixed - recognition
M2 $y'=e^t/(1+y^2)$
Detailed solution
Mixed - recognition
M3 $y''+6y'+9y=t$
Detailed solution
Mixed - recognition
M4 $\mathbf{x}'=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\mathbf{x}$
Detailed solution
Mixed - qualitative
M5 $y'=(y-1)(y+2)$
Detailed solution
Mixed - verification
M6 A student claims $y=1/(1-t)$ solves $y'=y$ , $y(0)=1$ . Diagnose the first mismatch.
Detailed solution

Round A checkpoint

Your six first moves should include linear integrating factor, separation, homogeneous-plus-particular response, phase-plane eigenanalysis, autonomous phase line, and substitution.

Round B: execute and verify

Mixed - first order
M7 Solve $y'=(1+t)y^2$ , $y(0)=-1$ . State the interval around zero on which the denominator remains nonzero.
Detailed solution
Mixed - first order
M8 Solve $y'+(2/t)y=t^2$ , $t\gt 0$ , $y(1)=1$ .
Detailed solution
Mixed - second order
M9 Solve $y''+2y'+2y=0$ , $y(0)=1$ , $y'(0)=0$ . Describe its phase portrait.
Detailed solution
Mixed - forcing
M10 Find the general solution of $y''+y=\sin3t$ .
Detailed solution
Mixed - numerical
M11 Use two Euler steps of size $0.5$ for $y'=y(2-y)$ , $y(0)=0.5$ . Check the values against the phase line.
Detailed solution
Mixed - system
M12 Classify the origin for $\mathbf{x}'=\begin{bmatrix}-2&1\\1&-2\end{bmatrix}\mathbf{x}$ and identify the late-time direction for generic initial data.
Detailed solution

Round C: challenge transfers

C1. Design a safe infusion model

A medication concentration $C(t)$ in a fixed blood volume $V$ receives a constant infusion $I$ mg/hour and clears at a rate proportional to concentration, $kC$ mg/hour.

Build the differential equation with units.
Find and classify the equilibrium concentration.
Solve for $C(0)=0$ .
Express the time to reach 90% of equilibrium.
Explain which parameters change the target and which change only the approach speed.

Detailed solution

C2. Compare two models, not two formulas

Two populations start at $P(0)=1$ :

$P_1'=P_1,\qquad P_2'=P_2(1-P_2/10).$

Compare them through rate laws, direction fields, early-time behavior, long-run behavior, and numerical sensitivity. Find the first nonzero term at which their Taylor expansions around $t=0$ differ.

Detailed solution

C3. Reverse-engineer a portrait

Construct a 2x2 linear system with a spiral sink whose trajectories rotate counterclockwise and whose amplitude envelope is $e^{-0.5t}$ . Give one valid matrix, its eigenvalues, and the corresponding scalar second-order equation for the first component.

Detailed solution

A four-session spacing plan

Do not finish the whole studio in one sitting.

Session	New work	Closed-book retrieval
Day 1	M1-M6	method map; equilibrium safeguard
Day 3	M7-M9	integrating-factor derivation; root cases
Day 7	M10-M12	phase-line and phase-plane classifications
Day 14	C1-C3	one worked example reconstructed from memory

After Day 14, create six new problems by changing coefficients or initial conditions. Mix them before solving so the labels disappear again.

Exit interview

Answer in complete sentences:

What do you inspect before choosing a method?
Which visual representation would you use for one autonomous state? For two coupled states?
How do you distinguish transient and forced behavior?
What evidence makes a numerical result credible?
Which error recurred in your log, and what cue will interrupt it?

I completed the studio across multiple sessions and wrote cue-specific corrections for my errors.

Next: Study the detailed solutions by comparing first decisions, not by copying final lines.

Detailed solutions

Compare reasoning, not handwriting

Use these solutions after an honest attempt. Put your work beside the solution and locate the first line where the decisions differ. A different valid method is not an error; explain why it works and compare its cost.

Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Mixed studio

Chapter 1

Solution 1.1

Rain contributes $+12$ m³/hour. Leakage contributes $-0.04W$ m³/hour because $0.04$ has units 1/hour and $W$ has units m³. Therefore

$W'=12-0.04W.$

Both right-side terms and $W'$ have units m³/hour. The zero-rate volume is $W=300$ m³, which also gives a behavioral check.

Solution 1.2

For $y=2+3e^{-t}$ , differentiate: $y'=-3e^{-t}$ . Substitute into the right side: $2-y=2-(2+3e^{-t})=-3e^{-t}=y'$ . The condition is $y(0)=2+3=5$ . The curve decreases toward the stable level 2, matching the sign of $2-y$ above 2.

Solution 1.3

The rate factors as $0.4N(1-N/50)$ . At $N=20$ , both factors are positive, so the state rises. At $N=50$ , the second factor is zero, so 50 is an equilibrium. At $N=80$ , the second factor is negative while $N$ is positive, so the state falls. The equation points positive populations toward 50.

Solution 1.4

Salt enters at $(3\text{ L/min})(0.08\text{ kg/L})=0.24$ kg/min. It leaves at $(3\text{ L/min})(S/200\text{ kg/L})=0.015S$ kg/min. Thus

$S'=0.24-0.015S,\qquad S(0)=10.$

At zero rate, $0.24=0.015S$ , so the model suggests a long-run amount of $S=16$ kg.

Solution 1.5

The coefficient $0.2$ has units $1/\text{hour}$ , so $D'(0)=-0.2(10)=-2$ g/hour.
Differentiate the proposed function: $D'=-2e^{-0.2t}=-0.2D$ . Also $D(0)=10$ , and its positive decreasing graph matches removal.
After one hour, $D(1)=10e^{-0.2}\approx8.1873$ g. The fraction lost is $1-e^{-0.2}\approx0.18127$ , or about 18.13%, not 20%.
Exact 20% loss at each hourly checkpoint instead gives $D_n=10(0.8)^n$ . A continuous exponential matching those checkpoints would use $D'=(\ln0.8)D\approx-0.22314D$ . An instantaneous percentage rate and a finite-interval percentage loss are different statements.

Solution 1.6

Net volume inflow is $2-1=1$ L/min, so $V(t)=100+t$ L for $t\ge0$ , as long as the tank does not overflow.
Salt enters at $(2\text{ L/min})(3\text{ g/L})=6$ g/min.
Well-mixed outflow has concentration $S(t)/(100+t)$ g/L. At 1 L/min, it removes $S(t)/(100+t)$ g/min. Therefore

$S'=6-\frac{S}{100+t},\qquad S(0)=0.$

Initially, $S'(0)=6$ g/min because there is no salt to remove yet. Both the amount and volume vary, so replacing the outflow concentration by $S/100$ would use the wrong volume. Exercise 5.6 solves this model.

Chapter 2

Solution 2.1

The equilibrium is $y=3$ . Below 3, $y-3\lt 0$ , so solutions fall; above 3, they rise. Arrows point away from 3, so it is unstable. The initial values 1 and 5 move down and up, respectively, while 3 remains constant.

Solution 2.2

For $y'=\cos t$ , each vertical column has a common slope because only time matters. For $y'=\cos y$ , each horizontal row has a common slope because only state matters. For $y'=\cos(t+y)$ , equal slopes lie on diagonal bands where $t+y$ is constant.

Solution 2.3

Between 0 and 1, $y'=y(1-y)\gt 0$ , so the solution rises toward the stable equilibrium 1. Differentiate along a solution:

$y''=y'(1-2y).$

While $0\lt y\lt 1/2$ , both factors are positive, so the graph is concave up. After $y$ passes $1/2$ , $1-2y\lt 0$ , so it is concave down. The growth rate peaks at the inflection state $y=1/2$ .

Solution 2.4

For $y=0$ , both $y'$ and $\sqrt y$ are zero. For $y=t^2/4$ with $t\ge0$ , $y'=t/2$ and $\sqrt y=t/2$ . Both satisfy $y(0)=0$ . The usual no-crossing/unique-future rule fails because $\sqrt y$ is not Lipschitz in $y$ at zero. In fact, solutions may wait at zero before departing.

Solution 2.5

The slope vanishes where $t-y=0$ , or $y=t$ . Below this line the rate is positive; above it the rate is negative. But the candidate $y=t$ has derivative 1 while the equation demands 0 along the line, so the zero-slope line is not a solution.

At $(0,0)$ , $y'(0)=0$ . Differentiating the equation along the solution gives $y''=1-y'$ , hence $y''(0)=1$ . The curve has a local minimum at zero and rises for sufficiently small positive time. As an optional check, $y=t-1+e^{-t}$ has the correct initial condition and derivative, with expansion $y=t^2/2+O(t^3)$ .

Solution 2.6

Differentiate along a solution, remembering that $y$ also depends on $t$ :

$y''=2t-y'=2t-t^2+y.$

At $(1,1)$ , $y'=0$ and $y''=2\gt 0$ , so this solution has a strict local minimum at $t=1$ . At $(-1,1)$ , $y'=0$ and $y''=-2\lt 0$ , so that solution has a strict local maximum at $t=-1$ . The smooth rate rule guarantees local solutions through both points. A horizontal field mark reports only the first derivative; the second derivative distinguishes these two turning behaviors.

Chapter 3

Solution 3.1

(a) $y'=t^2/y$ is separable: $y,dy=t^2dt$ . (b) $y'+e^t y=1$ is first-order linear with $p=e^t$ . (c) $y'=y\sin y$ is autonomous and separable; draw the phase line first. (d) $y''+9y=0$ is homogeneous second-order constant coefficient, so use characteristic roots.

Solution 3.2

Factor $y'=ty+t=t(y+1)$ , so $dy/(y+1)=t,dt$ : it is separable. The equation $y'=ty+1$ gives $y'-ty=1$ , but the right side $ty+1$ does not factor into a time-only and state-only product. Both equations are first-order linear; separation is the shorter option for the first.

Solution 3.3

On $t\gt 0$ , $y'=y/t$ is separable because $dy/y=dt/t$ . It is also homogeneous first-order linear: $y'-(1/t)y=0$ . Separation uses less setup. Both produce $y=Ct$ .

Solution 3.4

The sum $y^2+t$ cannot be factored as $a(t)b(y)$ , so it is not separable. The squared state violates the linear form $y'+p(t)y=q(t)$ . Useful next steps include drawing a direction field, computing a numerical solution for specified initial data, deriving local Taylor coefficients, and proving qualitative bounds or monotonicity.

Solution 3.5

For $t\ne2$ , divide by $t-2$ :

$y'+\frac{3}{t-2}y=t-2.$

This is first-order linear with $p(t)=3/(t-2)$ and $q(t)=t-2$ . The maximal coefficient-continuity interval containing 3 is $(2,\infty)$ ; the other interval is $(-\infty,2)$ . The next move would be an integrating factor, $\mu=|t-2|^3$ .

The original equation still makes an algebraic demand at $t=2$ : $3y(2)=0$ , so a differentiable extension must have $y(2)=0$ . This is a necessary condition, not a guarantee that every branch extends. Dividing by a coefficient that can vanish requires an interval check.

Solution 3.6

The state factor is $y(1+ty)$ , whose second factor mixes time and state; the $y^2$ term also rules out linearity in $y$ . First record the equilibrium $y=0$ . On a branch where $y\ne0$ , set $v=1/y$ . The chain rule gives

$v'=-\frac{y'}{y^2}=-\frac{y+ty^2}{y^2}=-v-t,$

so $v'+v=-t$ , a first-order linear equation with integrating factor $e^t$ . Linearity depends on the unknown: the transformed equation is linear in $v$ , while the original remains nonlinear in $y$ . Recovering $y=1/v$ requires $v\ne0$ , and the separate equilibrium $y=0$ cannot be recovered by taking a finite reciprocal.

Chapter 4

Solution 4.1

Separate and integrate:

$y,dy=3t^2dt,\qquad \frac{y^2}{2}=t^3+C.$

The condition gives $C=2$ , so $y^2=2t^3+4$ . Since $y(0)=2\gt 0$ , select

$y=\sqrt{2t^3+4}.$

Differentiation gives $y'=3t^2/y$ . The real solution interval containing zero requires $2t^3+4\gt 0$ , so $t\gt -\sqrt[3]{2}$ .

Solution 4.2

The division safeguard records equilibria $y=0$ and $y=2$ . For other solutions,

$\int\frac{dy}{y(y-2)}=t+C \quad\Rightarrow\quad \frac12\ln\left|\frac{y-2}{y}\right|=t+C.$

Exponentiating with a signed nonzero constant gives $(y-2)/y=Ce^{2t}$ , hence the non-equilibrium family $y=2/(1-Ce^{2t})$ , $C\ne0$ , on any interval where the denominator is nonzero. The equilibria $y=0$ and $y=2$ complete the solution set. Using $y(0)=1$ gives $C=-1$ , hence

$y=\frac{2}{1+e^{2t}}.$

This solution remains between the two equilibria and decreases toward zero, matching the phase line.

Solution 4.3

The equilibria are 0 and 500. Partial fractions or the standard logistic integration gives

$P=\frac{500}{1+Ae^{-0.6t}}.$

From $P(0)=50$ , $50=500/(1+A)$ , so $A=9$ :

$P(t)=\frac{500}{1+9e^{-0.6t}}.$

The rate is positive between 0 and 500 and negative above 500, so positive nonzero solutions approach 500.

Solution 4.4

Integrating $dy/(1+y^2)=dt$ gives $\arctan y=t+C$ . The condition $y(0)=1$ gives $C=\pi/4$ , so

$y=\tan(t+\pi/4).$

The nearest tangent poles occur when $t+\pi/4=-\pi/2$ and $t+\pi/4=\pi/2$ . The maximal interval containing zero is

$(-3\pi/4,\ \pi/4).$

Solution 4.5

Record $y=0$ before division; it does not satisfy the initial condition. For the negative branch, separate and integrate:

$\frac{dy}{y^2}=2t,dt,\qquad -\frac1y=t^2+C.$

Using $y(0)=-2$ gives $C=1/2$ , hence

$y=-\frac{2}{1+2t^2}.$

The denominator is positive for every real $t$ , so the maximal interval is $(-\infty,\infty)$ . Differentiation gives $y'=8t/(1+2t^2)^2=2ty^2$ , and $y(0)=-2$ . The curve decreases before zero and increases after zero, but never crosses $y=0$ : a change in derivative sign changes direction, not necessarily the sign of the state.

Solution 4.6

The equilibrium $y=0$ is excluded by $y(0)=4$ . On the positive branch,

$\int y^{-3/2},dy=\int dt \quad\Rightarrow\quad -\frac2{\sqrt y}=t+C.$

The initial condition gives $C=-1$ , so $2/\sqrt y=1-t$ . The left side is positive, which requires $t\lt 1$ . Therefore

$y=\frac4{(1-t)^2},\qquad -\infty<t<1.$

On this interval, $y'=8/(1-t)^3=y^{3/2}$ , and $y(0)=4$ . The solution blows up as $t\to1^-$ . Although the rational expression is also finite for $t\gt 1$ , there its derivative is negative while $y^{3/2}$ is positive. Squaring without preserving the branch inequality would introduce a false continuation.

Chapter 5

Solution 5.1

The zero-rate state in $y'=3y+6=3(y+2)$ is $-2$ , and arrows point away, so it is unstable. The integrating factor is $e^{-3t}$ , giving

$y=-2+Ce^{3t}.$

From $y(0)=1$ , $C=3$ , so $y=-2+3e^{3t}$ . It grows away from $-2$ , as predicted.

Solution 5.2

Here $p(t)=2/t$ , so on $t\gt 0$ , $\mu=t^2$ . Then

$(t^2y)'=t^3.$

Integration gives $t^2y=t^4/4+C$ , or $y=t^2/4+C/t^2$ . The condition $y(1)=0$ gives $C=-1/4$ :

$y=\frac{t^2}{4}-\frac{1}{4t^2}.$

Solution 5.3

Rewrite as $T'+0.25T=5$ , or read the equilibrium directly as 20. The solution is

$T=20+60e^{-0.25t}.$

Set $T=30$ : $10=60e^{-0.25t}$ , so $e^{-0.25t}=1/6$ and

$t=4\ln6\approx7.17\text{ time units}.$

Solution 5.4

The transient $Ce^{-t}$ is annihilated by the operator $D+1$ . Apply it to the proposed family:

$y'+y=(2t-Ce^{-t})+(t^2+Ce^{-t})=t^2+2t.$

Thus one valid equation is

$y'+y=t^2+2t.$

The polynomial $t^2$ is the forced response; $Ce^{-t}$ is the decaying transient.

Solution 5.5

The integrating factor is $e^{2t}$ , so

$(e^{2t}y)'=e^t,\qquad e^{2t}y=e^t+C.$

Thus $y=e^{-t}+Ce^{-2t}$ . The initial condition gives $C=-1$ , hence $y=e^{-t}-e^{-2t}$ . Its derivative is $-e^{-t}+2e^{-2t}$ , which is positive before $t=\ln2$ and negative afterward. On $t\ge0$ , the maximum is therefore

$y(\ln2)=\frac12-\frac14=\frac14.$

Finally, $y'+2y=e^{-t}$ , and $y(0)=0$ . The solution first builds up under the input, then falls to zero as the input fades.

Solution 5.6

On $t\ge0$ , the integrating factor is $\mu=100+t$ . Multiplying every term gives

$[(100+t)S]'=6(100+t).$

Integration yields $(100+t)S=3(100+t)^2+C$ . At $t=0$ , $S=0$ , so $C=-30000$ , and

$S=3(100+t)-\frac{30000}{100+t}.$

Consequently the concentration is

$\frac{S}{100+t}=3-\frac{30000}{(100+t)^2}\longrightarrow3\text{ g/L}.$

The concentration approaches that of the inflow, but volume grows without bound in this idealized no-overflow model, so the amount grows without a finite limit. As a check, $S'=3+30000/(100+t)^2=6-S/(100+t)$ , and $S'(0)=6$ g/min, as predicted in Exercise 1.6.

Chapter 6

Solution 6.1

The equilibria are 0, 1, and 3. The sign pattern of $y(y-1)(y-3)$ , from left to right, is $-,+,-,+$ . Arrows point away from 0, toward 1, and away from 3. Thus 0 and 3 are unstable, while 1 is stable.

Solution 6.2

The phase line gives: $y(0)=-1$ decreases away below 0; $y(0)=1/2$ rises toward 1; $y(0)=2$ falls toward 1; and $y(0)=4$ rises away above 3. The two inner solutions tend to 1 as $t\to\infty$ . The outer solutions instead become unbounded in finite positive time: separation expresses the time to reach a state as an integral of $1/[y(y-1)(y-3)]$ , and the integral to either infinite endpoint converges because its magnitude behaves like $1/|y|^3$ . Thus their arrows must not be interpreted as solutions that remain finite for every positive time.

Solution 6.3

Treat $P'=rP-(r/K)P^2$ as a function of state. Differentiate with respect to $P$ :

$\frac{dP'}{dP}=r-\frac{2r}{K}P.$

With $r,K\gt 0$ , the critical state is $P=K/2$ , and the downward-opening parabola makes it a maximum. The maximum rate is $rK/4$ . The carrying-capacity population is the equilibrium $K$ ; these have different units. A population may start above $K$ , but then its rate is negative, so $K$ is not an absolute upper bound on every initial state.

Solution 6.4

For $a\gt 0$ , equilibria are $\pm\sqrt a$ . The rate is positive outside them and negative between them, so $-\sqrt a$ is stable and $+\sqrt a$ is unstable. For $a=0$ , $y'=y^2\ge0$ ; zero is semistable, attracting from below and repelling above. For $a\lt 0$ , $y'=y^2+|a|\gt 0$ , so no equilibria exist and every solution rises on its interval of existence.

Solution 6.5

The only equilibrium is $y=2$ . Below it, $(y-2)^3\lt 0$ , so the rate is positive. Above it, the rate is negative. Both arrows point inward: 2 is asymptotically stable.

Yet $f'(y)=-3(y-2)^2$ , so $f'(2)=0$ . A zero derivative makes the first-derivative test inconclusive, not unstable. Starting at 1, the solution rises toward 2; starting at 3, it falls toward 2. Smoothness prevents crossing the equilibrium. Each trajectory is monotone and bounded, and any limit different from 2 would retain a nonzero rate, so both limits must equal 2.

Solution 6.6

Complete the square in the rate:

$P'=2-H-\frac{(P-4)^2}{8}.$

For $0\le H\lt 2$ , the two equilibria are

$P_-=4-\sqrt{16-8H},\qquad P_+=4+\sqrt{16-8H}.$

The rate is positive between them and negative outside. Thus $P_-$ is unstable and $P_+$ is asymptotically stable. For $0\lt H\lt 2$ , a starting population between 0 and $P_-$ declines to extinction, while a starting population above $P_-$ approaches $P_+$ . The lower equilibrium is a survival threshold. At $H=0$ , it is zero itself, and positive populations instead approach 8.

At $H=2$ , $P'=-(P-4)^2/8$ : the double equilibrium 4 attracts from above and repels from below, so it is semistable. At $H\gt 2$ , the rate is strictly negative everywhere and there are no equilibria; positive populations reach extinction. Because a constant harvest would otherwise drive $P$ below zero, the physical model stops at extinction. The threshold $H=2$ equals the unharvested model's maximum growth rate.

Chapter 7

Solution 7.1

The characteristic equation $r^2-r-6=(r-3)(r+2)=0$ gives roots 3 and $-2$ . Thus

$y=C_1e^{3t}+C_2e^{-2t}.$

The conditions give $C_1+C_2=1$ and $3C_1-2C_2=0$ , hence $C_1=2/5$ , $C_2=3/5$ . The growing $e^{3t}$ mode dominates for large positive time.

Solution 7.2

The characteristic polynomial is $(r+3)^2$ , so

$y=(C_1+C_2t)e^{-3t}.$

The first condition gives $C_1=0$ . Differentiating then evaluating at zero gives $C_2=2$ . Therefore $y=2te^{-3t}$ .

Solution 7.3

The roots are $-\zeta\pm\sqrt{\zeta^2-1}$ . At $\zeta=0$ , they are pure imaginary and the phase portrait is a center. At $1/2$ , they are complex with negative real part, producing a spiral sink. At 1, the negative root repeats, giving critical damping and a degenerate stable node. At 2, both roots are real and negative, giving an overdamped stable node.

Solution 7.4

Envelope $e^{-t}$ requires real part $-1$ . Period $\pi$ requires angular frequency $2$ , since $2\pi/\beta=\pi$ . Desired roots are $-1\pm2i$ . Their polynomial is

$(r+1)^2+4=r^2+2r+5.$

Choose $a=2$ , $b=5$ .

Solution 7.5

The roots are $\pm2i$ , so $x=C_1\cos2t+C_2\sin2t$ . Position gives $C_1=1$ , and velocity gives $2C_2=2$ , hence

$x=\cos2t+\sin2t=\sqrt2\cos(2t-\pi/4).$

The amplitude is $\sqrt2$ and the period is $2\pi/2=\pi$ . For $E=(x')^2/2+2x^2$ , differentiate:

$E'=x'x''+4xx'=x'(x''+4x)=0.$

Initially $E=2^2/2+2(1)^2=4$ , so $E=4$ at every time. This gives an independent physical check: the undamped oscillator exchanges kinetic and potential energy without losing their sum.

Solution 7.6

The characteristic equation is $r(r+2)=0$ . Therefore

$y=C_1+C_2e^{-2t},\qquad y'=-2C_2e^{-2t}.$

Velocity gives $-2C_2=-4$ , so $C_2=2$ . Position then gives $C_1=-1$ , hence $y=-1+2e^{-2t}$ and $\lim_{t\to\infty}y=-1$ . Direct substitution gives $y''+2y'=8e^{-2t}-8e^{-2t}=0$ .

The root 0 supplies the constant mode $e^{0t}=1$ , while the root $-2$ supplies a decaying mode. Undamped oscillation needs a nonzero imaginary pair; a real zero root alone gives no rotation or periodic motion.

Chapter 8

Solution 8.1

Natural roots are $\pm i$ , so $y_h=C_1\cos t+C_2\sin t$ . For the polynomial input $t$ , try $y_p=At+B$ . Since $y_p''=0$ , matching $At+B=t$ gives $A=1,B=0$ . Thus

$y=C_1\cos t+C_2\sin t+t.$

Solution 8.2

The homogeneous solution is $C_1\cos2t+C_2\sin2t$ . Try $y_p=A\cos t+B\sin t$ ; substitution gives $3A=1$ , $3B=0$ . Thus $y_p=\frac13\cos t$ . The conditions give $C_1=-1/3$ , $C_2=0$ , so

$y=\frac{\cos t-\cos2t}{3}.$

Solution 8.3

(a) $Ae^{-t}$ ; it does not collide with the natural modes. (b) $t(A\cos2t+B\sin2t)$ ; frequency 2 is the natural frequency, so multiply once by $t$ . (c) $At^2+Bt+C$ ; use the complete polynomial family.

Solution 8.4

For $y''+\omega_0^2y=F\cos(\omega t)$ with zero initial position and velocity, $\omega_0\gt 0$ , and $\omega\ne\omega_0$ , the response is

$y=\frac{F[\cos(\omega t)-\cos(\omega_0t)]}{\omega_0^2-\omega^2}.$

The identity rewrites the numerator as a fast oscillation with average angular frequency $(\omega+\omega_0)/2$ , multiplied by $-2\sin((\omega-\omega_0)t/2)$ . Close frequencies produce a slowly varying amplitude, or beats. In the limit $\omega\to\omega_0$ , both numerator and denominator vanish; their ratio tends to $Ft\sin(\omega_0t)/(2\omega_0)$ . It is this full response, not the numerator alone, whose resonant envelope grows linearly.

Solution 8.5

The homogeneous characteristic polynomial is $(r-1)^2$ , so both $e^t$ and $te^t$ are natural modes. The forcing trial must therefore be $At^2e^t$ , not $Ae^t$ or $Ate^t$ .

For a direct verification, set $y=e^t u$ . Then

$y''-2y'+y=e^t u''.$

The equation becomes $u''=1$ , so $u=t^2/2+C_1+C_2t$ . Since $y(0)=u(0)=0$ , $C_1=0$ . Also $y'(0)=u(0)+u'(0)=0$ , so $C_2=0$ . Thus

$y=\frac12t^2e^t.$

The transformed equation verifies the forcing, and evaluation verifies both conditions. The homogeneous modes grow rather than decay, so they should not be described as decaying transients.

Solution 8.6

The roots of $r^2+2r+2=0$ are $-1\pm i$ , giving $y_h=e^{-t}(C_1\cos t+C_2\sin t)$ . Try $y_p=A\cos t+B\sin t$ . Substitution yields

$(A+2B)\cos t+(B-2A)\sin t=\cos t.$

Thus $A+2B=1$ , $B-2A=0$ , so $A=1/5$ , $B=2/5$ . Apply the initial conditions to the full solution: $C_1+1/5=0$ and $-C_1+C_2+2/5=0$ . Therefore

$y=e^{-t}\left(-\frac15\cos t-\frac35\sin t\right) +\frac15\cos t+\frac25\sin t.$

The steady-state amplitude is $\sqrt{(1/5)^2+(2/5)^2}=1/\sqrt5$ . The first term is the decaying transient. The difference of any two solutions satisfies the homogeneous equation, whose modes decay, so their periodic responses agree asymptotically.

Chapter 9

Solution 9.1

The system is already diagonal:

$x=C_1e^{-t},\qquad y=C_2e^{-3t}.$

Both modes decay, so the origin is a stable node. Unless $C_1=0$ , the slower $e^{-t}$ mode dominates late, and trajectories approach tangent to the $x$ -axis.

Solution 9.2

The matrix has eigenvalue 3 with eigenvector $[1,1]^T$ , and eigenvalue 1 with eigenvector $[1,-1]^T$ . Therefore

$\mathbf{x}=C_1e^{3t}\begin{bmatrix}1\1\end{bmatrix}+C_2e^t\begin{bmatrix}1\-1\end{bmatrix}.$

Both modes grow, so the origin is an unstable node; the $[1,1]^T$ direction dominates generically.

Solution 9.3

Let $x_1=y$ , $x_2=y'$ . Then

$\begin{bmatrix}x_1'\x_2'\end{bmatrix} =\begin{bmatrix}0&1\-5&-2\end{bmatrix} \begin{bmatrix}x_1\x_2\end{bmatrix}.$

The eigenvalues solve $\lambda^2+2\lambda+5=0$ , giving $-1\pm2i$ . The negative real part and nonzero imaginary part produce a spiral sink.

Solution 9.4

The eigenvalue discriminant is $D=\tau^2-4\Delta$ . If $\Delta\lt 0$ , eigenvalues have opposite signs: saddle. If $\Delta\gt 0$ and $D\gt 0$ , the roots are real with common sign set by $\tau$ : stable node for $\tau\lt 0$ , unstable node for $\tau\gt 0$ . If $D\lt 0$ , they are complex: spiral sink for $\tau\lt 0$ , spiral source for $\tau\gt 0$ , and center in the ideal linear case $\tau=0$ . The curves $\Delta=0$ and $D=0$ are classification boundaries.

On $D=0$ with $\Delta\gt 0$ , the repeated nonzero eigenvalue is $\tau/2$ : its sign still determines decay or growth, but eigenvectors distinguish a scalar-matrix node from a defective node. On $\Delta=0$ , at least one eigenvalue is zero and the origin is not an isolated equilibrium. If $\tau\lt 0$ , trajectories approach a line of equilibria, so the origin is stable but not asymptotically stable; if $\tau\gt 0$ , it is unstable. At $(\tau,\Delta)=(0,0)$ , $A=0$ leaves every point fixed, whereas a nonzero nilpotent matrix permits linear drift. Trace and determinant alone cannot distinguish these last two cases.

Solution 9.5

First solve $y'=-y$ , $y(0)=1$ , obtaining $y=e^{-t}$ . Then $x'+x=e^{-t}$ , so the integrating factor $e^t$ gives $(e^tx)'=1$ . Using $x(0)=0$ yields

$x=te^{-t},\qquad y=e^{-t}.$

Differentiate to check: $x'=(1-t)e^{-t}=-x+y$ , and $y'=-e^{-t}=-y$ .

The matrix $\begin{bmatrix}-1&1\0&-1\end{bmatrix}$ has repeated eigenvalue $-1$ , but its eigenvectors satisfy $v_2=0$ , leaving only the $x$ -axis direction. An eigenvector mode has zero second component and cannot match $y(0)=1$ . The additional $te^{-t}$ contribution is essential. In general, $y=C_2e^{-t}$ and $x=(C_1+C_2t)e^{-t}$ ; every solution tends to zero, so the origin is a defective stable node.

Solution 9.6

Differentiate $x'=-4y$ to get $x''=-4y'=-4x$ . Since $x(0)=2$ and $x'(0)=0$ ,

$x=2\cos2t,\qquad y=-x'/4=\sin2t.$

You can also prove the invariant without solving:

$\frac{d}{dt}(x^2+4y^2)=2x(-4y)+8yx=0.$

Its initial value is 4, so the orbit is the ellipse $x^2+4y^2=4$ , or $x^2/4+y^2=1$ . At $(2,0)$ , the velocity is $(0,2)$ , pointing upward, so rotation is counterclockwise. The period is $\pi$ . The origin is a center: nearby trajectories remain nearby, but nonzero invariant level sets never approach it. Thus it is stable, not attracting or asymptotically stable.

Chapter 10

Solution 10.1

With $h=0.2$ , use $y_{n+1}=y_n+0.2(y_n-t_n)$ :

$t_n$	$y_n$	slope	next value
0.0	1.0	1.0	1.2
0.2	1.2	1.0	1.4
0.4	1.4	1.0	1.6

Thus $y(0.6)\approx1.6$ .

In fact, $y=1+t$ satisfies both the equation and initial condition, so Euler is exact at these grid points in exact arithmetic. This special straight-line solution is not evidence that Euler generally has zero error.

Solution 10.2

Euler gives $y_{n+1}=(1+h)y_n$ , so from $y_0=1$ , $y_n=(1+h)^n$ . At $t=1$ , $n=1/h$ :

$h=1$ : $2$ ;
$h=1/2$ : $1.5^2=2.25$ ;
$h=1/4$ : $1.25^4\approx2.4414$ .

The values rise toward $e\approx2.7183$ as the step shrinks.

Solution 10.3

The phase line shows that solutions starting between 0 and 10 rise toward 10 without crossing it. A useful numerical test should therefore: assert $0\lt P_n\lt 10$ , rerun with half the step, compare the two trajectories over the same times, and reject a method or step size that crosses the invariant equilibrium. A direction-field overlay supplies an independent visual check.

Solution 10.4

Euler gives $y_{n+1}=(1-ah)y_n$ . Decay in magnitude requires

$|1-ah|<1\quad\Rightarrow\quad 0<h<\frac{2}{a}.$

Decay without alternating sign requires $0\le1-ah\lt 1$ , hence

$0<h\le\frac1a.$

At $h=1/a$ , the numerical state reaches zero in one step.

Solution 10.5

For $f(t,y)=y-t$ , the midpoint stages are:

Start $(t_n,y_n)$	$k_1$	Midpoint $(t_n+h/2,y_n+hk_1/2)$	$k_2$	Next $y$
$(0,0)$	0	$(0.25,0)$	-0.25	-0.125
$(0.5,-0.125)$	-0.625	$(0.75,-0.28125)$	-1.03125	-0.640625

Euler instead gives $y_1=0$ and $y_2=0+0.5(0-0.5)=-0.25$ . The exact value is $y(1)=2-e\approx-0.718282$ , making the absolute errors about 0.077657 for midpoint and 0.468282 for Euler.

The midpoint result is closer in this comparison because it samples the rate inside each step. It uses four rate evaluations rather than Euler's two, so the comparison fixes step size, not computational cost. One example does not prove a convergence order; repeated step-halving tests provide that evidence.

Solution 10.6

Write $a=hr$ . For $0\lt z_n\lt1$ and $0\lt a\le1$ , the update $z_{n+1}=z_n+az_n(1-z_n)$ is positive and greater than $z_n$ . For its upper bound, factor

$1-z_{n+1}=(1-z_n)(1-az_n).$

Both factors are positive because $z_n\lt 1$ and $az_n\lt 1$ . Thus $z_{n+1}\lt 1$ , and induction preserves the interval for every finite step in exact arithmetic.

For $r=1$ , $h=2$ , and $z_0=0.75$ ,

$z_1=0.75+2(0.75)(0.25)=1.125,$

which crosses the equilibrium even though the exact solution cannot. The bound $hr\le1$ is sufficient to preserve the interval for every starting state within it; it does not estimate the time-trajectory error. Step-halving comparisons are still needed for accuracy, and floating-point implementations should allow roundoff-sized tolerances.

Mixed studio

Solution M1

Rewrite as $y'-ty=\sin t$ . It is first-order linear with $p(t)=-t$ , so $\mu=e^{-t^2/2}$ . A direction field or substitution can check the eventual solution.

Solution M2

The factors separate: $(1+y^2)dy=e^t dt$ . Integration gives $y+y^3/3=e^t+C$ , generally best left implicit unless an initial condition selects a useful form.

Solution M3

Use homogeneous plus particular response. The characteristic root $-3$ repeats, so $y_h=(C_1+C_2t)e^{-3t}$ . For the polynomial input, try $y_p=At+B$ .

Solution M4

The matrix has eigenvalues $\pm i$ . The system is $x'=-y, y'=x$ , so $d(x^2+y^2)/dt=0$ . Trajectories are counterclockwise circles centered at the origin, an ideal center.

Solution M5

Use an autonomous phase line before separation. Equilibria are $-2$ and 1. The sign pattern is positive outside and negative between, so $-2$ is stable and 1 unstable.

Solution M6

The condition holds, but the equation does not. The derivative is $y'=1/(1-t)^2=y^2$ , not $y$ . The claimed function solves $y'=y^2$ instead.

Solution M7

The equilibrium $y=0$ does not match the condition. Separate:

$\int y^{-2}dy=\int(1+t)dt \quad\Rightarrow\quad -\frac1y=t+\frac{t^2}{2}+C.$

At $t=0,y=-1$ , $C=1$ , so

$y=-\frac{1}{1+t+t^2/2}.$

The quadratic denominator has discriminant $1-2\lt 0$ , so it never vanishes; the solution exists for all real $t$ .

Solution M8

On $t\gt 0$ , $\mu=t^2$ . Then $(t^2y)'=t^4$ , so $t^2y=t^5/5+C$ . The condition gives $1=1/5+C$ , hence $C=4/5$ :

$y=\frac{t^3}{5}+\frac{4}{5t^2}.$

Solution M9

Roots are $-1\pm i$ , so $y=e^{-t}(C_1\cos t+C_2\sin t)$ . The conditions give $C_1=1$ and $-C_1+C_2=0$ , so

$y=e^{-t}(\cos t+\sin t).$

Negative real part plus rotation gives a spiral sink in the $(y,y')$ phase plane.

Solution M10

Natural motion is $C_1\cos t+C_2\sin t$ . Try $A\sin3t+B\cos3t$ . Matching gives $-8A=1$ , $-8B=0$ , hence

$y=C_1\cos t+C_2\sin t-\frac18\sin3t.$

Solution M11

At $y_0=0.5$ , the slope is $0.5(1.5)=0.75$ , so $y_1=0.5+0.5(0.75)=0.875$ . The next slope is $0.875(1.125)=0.984375$ , so $y_2=1.3671875$ . Both values remain between the unstable equilibrium 0 and stable equilibrium 2 and move upward, matching the phase line.

Solution M12

The matrix has eigenvalue $-1$ along $[1,1]^T$ and $-3$ along $[1,-1]^T$ . Both decay, so the origin is a stable node. The slower $e^{-t}$ mode dominates generically, so trajectories approach tangent to $[1,1]^T$ .

Solution C1

If $C$ is concentration in mg/L, the amount in the fixed volume is $VC$ . Its balance is $V C'=I-kC$ , so

$C'+\frac{k}{V}C=\frac{I}{V}.$

Here $I$ has units mg/hour, $k$ has units L/hour, and every term in the amount balance has units mg/hour. The equilibrium is $C_$ , stable because the displacement equation is $(C-C_$ )'=-(k/V)(C-C_*) $(C - C_{*})^{'} = - (k / V) (C - C_{*})$ . From $C(0)=0$ ,

$C(t)=\frac{I}{k}\left(1-e^{-kt/V}\right).$

For 90% of equilibrium, $e^{-kt/V}=0.1$ , so $t=(V/k)\ln10$ . The ratio $I/k$ sets the target; $k/V$ sets the approach rate, though changing $k$ affects both.

Solution C2

Both models start with positive rate, but $P_1$ has unlimited positive feedback while $P_2$ weakens as it approaches 10. Thus $P_1=e^t$ grows without bound, while $P_2$ approaches 10. At $P=1$ , rates are 1 and 0.9, so their Taylor series already differ in the linear term:

$P_1=1+t+O(t^2),\qquad P_2=1+0.9t+O(t^2).$

The exponential field has horizontal-row slopes proportional to $P$ ; the logistic field adds a zero-slope row at 10 and reverses sign above it. Numerical checks for the logistic model should preserve the interval $(0,10)$ .

Solution C3

Counterclockwise rotation with envelope $e^{-0.5t}$ can use

$A=\begin{bmatrix}-0.5&-1\1&-0.5\end{bmatrix}.$

Its eigenvalues are $-0.5\pm i$ . Since $x'=-0.5x-y$ , differentiating and eliminating $y$ gives

$x''+x'+1.25x=0.$

The roots of this scalar equation are the same $-0.5\pm i$ , confirming the envelope and angular frequency.

I compared first decisions, updated my error log, and re-solved missed problems without looking.

Next: See the evidence and design choices behind this workbook.

Learning design and sources

Why the workbook works this way

This is an independent differential-equations workbook. It does not reproduce text, exercises, or artwork from David Klein's Organic Chemistry as a Second Language, and it is not affiliated with Klein or Wiley. It transfers a public, high-level teaching pattern: turn an intimidating subject into a small language of reusable skills, explain each move, and require the learner to practice immediately.

What the chemistry workbook establishes

Wiley describes Klein's book as a skill-building supplement that teaches learners to ask useful questions, connects critical principles across the course, places hands-on exercises and step-by-step explanations in each section, and moves from single-skill Practice Problems to Integrated Problems and Challenge Problems. Wiley also highlights Klein's use of analogy and an informal presentation. Those are documented features of the book, not proof of causal effectiveness. See the official Wiley description.

I could not locate a controlled study that isolates the workbook as an intervention. A Journal of Chemical Education review establishes contemporary scholarly attention, but a review is not an efficacy trial. Popularity, testimonials, and publisher language should not be converted into learning-effect claims.

The defensible conclusion is narrower: the public structure of the book aligns with several instructional practices supported by broader research. This workbook adopts those practices and makes the alignment explicit.

How the pattern transfers

Learning design	Workbook implementation	Learner action
Coherent language, not a bag of tricks	state-rate-solution vocabulary and one recurring four-question loop	say what the equation means before calculating
Recognition cues	fingerprint cards and mixed method trainer	name the structural cue that selects a method
Worked examples	numbered steps pair each operation with a reason	explain why each step is legal
Faded guidance	"I do / we do / you do" and backward-completion examples	supply progressively earlier steps
Immediate practice	Practice items sit beside each new method	attempt before opening a solution
Integrated and challenge work	every chapter ends with three difficulty levels	combine old and new skills
Detailed feedback	solution bank includes method choice, calculation, and behavior checks	find the first divergent decision
Visual analogy	fields, phase lines, time traces, and phase portraits show the same rule	translate among symbolic, graphical, numerical, and verbal forms

Why examples fade

Novices can spend working memory on unproductive search when they face an unfamiliar problem with no model. A major review of worked-example research describes how examples can help learners build reusable mental patterns, especially when examples integrate steps with explanations and prompt self-explanation (Atkinson et al., 2000).

The handoff to independent solving matters. Research comparing example-problem pairs and fading found benefits from both, with backward fading especially useful for intra-mathematical skills in the reported study (Große, 2015, ERIC record). The US What Works Clearinghouse likewise recommends interleaving worked examples with problem solving and combining graphics with verbal descriptions (IES practice guide).

That evidence motivates the recurring sequence:

see a complete example;
complete its missing late steps;
complete a similar problem with earlier steps removed; and
solve a mixed problem without a method label.

The book reduces guidance as the learner gains experience. This avoids turning a helpful scaffold into permanent dependence.

Why practice becomes mixed

Blocked practice, which groups problems by method, teaches execution while quietly giving away the method. Mixed practice forces a separate skill: discriminating among strategies from problem structure. In a cluster-randomized study summarized by the Institute of Education Sciences, classes receiving more interleaved, or mixed, mathematics practice outperformed the blocked-practice control on an unannounced delayed test. The two groups received the same problems in different orders (IES study summary).

This workbook therefore uses short blocked sets while a move is new, then mixes it with older moves. The method trainer, integrated problems, mixed studio, and four-session schedule all remove chapter-label cues.

Why every equation gets several pictures

Differential-equations students can manipulate a formula without treating a solution as a function or connecting a rate rule to graphical behavior. Research on student thinking in ordinary differential equations identifies this "function-as-solution" difficulty and the importance of students' intuitions and images (Rasmussen, 2001).

Direction-field instruction built around guided inquiry has been developed specifically to address students' conceptions of ODE solutions (Hyland, van Kampen, and Nolan, 2021). Later interview research found that students noticed and valued the intervention's emphasis on conceptual questioning and active interaction, though those perception data do not by themselves establish achievement gains (2023 ERIC record).

The interactive labs repeatedly coordinate:

an equation's symbols;
the local rate field;
a solution over time;
a phase line or phase portrait;
a numerical approximation; and
a sentence about feedback or motion.

The controls do not merely animate a finished graph. They ask the learner to move an initial condition, change a parameter, predict a transition, and then compare the result with the equation.

Why errors appear beside correct work

An incorrect answer becomes useful when the learner explains the first invalid decision. IES describes example-based mathematics activities that present correct and incorrect fictitious student work, target common misconceptions, and require explanation before a matched problem (IES overview of AlgebraByExample).

The error clinics in this book focus on high-leverage misconceptions:

dividing away equilibrium solutions;
treating a sum as separable;
reading $p(t)$ before normalizing a linear equation;
confusing a phase line with a time graph;
forgetting the second mode at a repeated root;
colliding a forced-response trial with natural motion; and
trusting a numerical plot without a convergence check.

Why retrieval repeats

The end of each chapter asks the learner to close or cover the page and reconstruct a small set of ideas. Later chapters retrieve earlier methods in a different context. The aim is durable access, not a feeling of familiarity produced by rereading.

The evidence base for retrieval and spacing extends beyond this specific book and subject. This workbook uses a conservative implementation: short closed-book prompts, immediate correction, and revisits after increasing gaps. It does not claim that one schedule fits every learner.

Design limitations

This book is a first-course workout, not a complete differential-equations text. It omits exact equations, Laplace transforms, power-series methods, boundary-value problems, partial differential equations, rigorous existence theory, and nonlinear systems beyond an introduction. It also cannot observe a learner's written reasoning or adapt problem difficulty automatically.

Interactive graphs support exploration, but hand sketching and algebra remain essential. A learner should use the canvases to test predictions, not as a substitute for making them.

A testable definition of effectiveness

The workbook succeeds only if a learner can do more than reproduce its examples. A useful evaluation would measure whether learners can:

classify unfamiliar equations without chapter labels;
translate among rate laws, fields, and solution behavior;
retain methods after a delay;
detect realistic erroneous solutions;
choose analytic or numerical tools appropriately; and
transfer the reasoning to a new model.

Those outcomes should be compared with a relevant alternative using delayed, mixed assessments. Until such a study exists, describe this workbook as evidence-informed, not evidence-proven.

Source selection favors official publisher descriptions, government evidence summaries, peer-reviewed research records, and original scholarly articles. Claims about a source's implications remain scoped to the population and design studied.

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Differential Equations as a Second Language