Learn the language of change

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:

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

The workout rhythm

Each chapter repeats the same deliberate sequence.

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

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.

No interactive lab sends data anywhere. Checkpoints stay in your browser. You can clear them from the button below.

A two-minute preflight

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

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

Suppose (T(t)) measures a cup of coffee's temperature in degrees Celsius, while (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;
• (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.

Units expose meaning

Every term added to another term must have the same units. In

$$T^{\prime }=-k(T-22),$$

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 dimensionless proportionality constant: its right side would have units °C² rather than °C/min.

Worked example: read a model from a story

Fade the support

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.

Verify in three passes

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

1. Differentiate: (y'=-12e^{-3t}).
2. Substitute: (-3y=-3(4e^{-3t})=-12e^{-3t}=y').
3. 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.

Practice

1. 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

2. Practice - verification

   1.2 Verify whether \(y=2+3e^{-t}\) solves \(y'=2-y\) and \(y(0)=5\). Use all three passes.

   Detailed solution

3. 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

4. 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

Retrieve before you leave

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

1. What are the four objects you name when reading a model?
2. Why does a differential equation usually have a family of solutions?
3. 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

Try these moves before reading on:

1. Choose (y'=-0.65y). Place the initial point above, on, and below (y=0).
2. Choose the logistic equation. Find every horizontal row where the marks go flat.
3. 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)), a horizontal zero-slope row often gives an equilibrium solution.

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

2. Mark the sign regions

• (y'>0): solution curves rise as (t) increases.
• (y'<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.

Worked example: predict without solving

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].

What can go wrong?

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

1. 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

2. 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

3. 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

4. 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

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.

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. Blocked homework hides that decision because every exercise under "Separable Equations" is separable. Real problems do not carry those labels.

This chapter builds a classifier you will retrieve and refine throughout the book.

Normalize, then inspect

Before naming a method:

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

The same equation can wear a disguise. For example,

$$(1+t)y^{\prime }=2y$$

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

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

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

The first-course method map

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

Contrast cases: train the boundary

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

Worked example: name two routes

Fade the classification

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

1. (y'=(t^2+1)e^{-y})
2. (y'-\frac{1}{t}y=t)
3. (y'=y^2-4)
4. (y''-5y'+6y=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."

Practice

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

1. 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

2. 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

3. Integrated - multiple routes

   3.3 Find every applicable early-course method for \(y'=y/t\), \(t>0\). Which route uses less machinery?

   Detailed solution

4. 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

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:

$$y^{\prime }=a(t)b(y).$$

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

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

Worked example: full support

Why the absolute value disappears carefully

From (\ln|y|=F(t)+C), exponentiation gives (|y|=Ae^{F(t)}) with (A>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).

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.

Implicit can be finished

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

$${\int }_1^{y(t)}(1+u^2)du={\int }_0^{t}sds,$$

so

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

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.

Error clinic

Practice

1. 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

2. 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

3. 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

4. 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

Spaced retrieval 2

Without notes:

1. rebuild the three-pass verification from Chapter 1;
2. state the horizontal-row fingerprint from Chapter 2; and
3. 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

$$y^{\prime }+p(t)y=q(t).$$

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

$$\mu y^{\prime }+\mu py=(\mu y{)}^{\prime }.$$

Derive the method once

The product rule says

$$(\mu y{)}^{\prime }=\mu y^{\prime }+{\mu }^{\prime }y.$$

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 reliable sequence

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

Worked example: full support

Backward-faded example

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

When linear and separable overlap

The homogeneous linear equation (y'+p(t)y=0) is also separable:

$$\frac{y^{\prime }}{y}=-p(t).$$

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.

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

$$S^{\prime }=\underset{\text{rate in}}{\underbrace{(2)(0.1)}}-\underset{\text{rate out}}{\underbrace{(2)\frac{S}{100}}}=0.2-0.02S.$$

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

Practice

1. 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

2. Practice - variable coefficient

   5.2 Solve \(y'+(2/t)y=t\) for \(t>0\), with \(y(1)=0\).

   Detailed solution

3. 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

4. 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

Retrieve and discriminate

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

1. (y'=t(1-y)) by separation;
2. (y'+ty=t) by an integrating factor; and
3. explain why the surface similarity does not make the methods interchangeable.

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:

$$y^{\prime }=f(y).$$

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

1. Solve (f(y)=0) for equilibrium states.
2. Use those states to divide the line into intervals.
3. Test the sign of (f(y)) on each interval.
4. Draw an arrow up for (f(y)>0) and down for (f(y)<0).
5. Classify each equilibrium by the arrows around it.

An equilibrium is:

• stable when nearby arrows point toward it;
• unstable when nearby arrows point away; and
• semistable when one side points in and the other points out.

Lab: logistic feedback

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

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}}).

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_*)<0) suggests local stability;
• (f'(y_*)>0) suggests local instability.

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

Parameters can change the portrait

Consider (y'=y(r-y)). When (r>0), (y=0) is unstable and (y=r) is stable. When (r<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

Practice

1. Practice - phase line

   6.1 Draw and classify the phase line for \(y'=y(y-1)(y-3)\).

   Detailed solution

2. 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

3. Integrated - logistic interpretation

   6.3 In \(P'=rP(1-P/K)\), show that the growth rate \(P'\) is largest at \(P=K/2\). Distinguish "largest growth rate" from "largest population."

   Detailed solution

4. Challenge - threshold

   6.4 Analyze \(y'=y^2-a\) for \(a>0\), \(a=0\), and \(a<0\). Track how the equilibria and their stability change.

   Detailed solution

Retrieve and connect

Without looking, answer:

1. Why must separation record equilibria before division?
2. How does an integrating factor use the product rule?
3. 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:

$$y^{\prime \prime }+a{y}^{\prime }+by=0,y(0)=y_0,y^{\prime }(0)=v_0.$$

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}):

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

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

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

The roots predict the motion's geometry.

Lab: watch damping reshape motion

Move (c) slowly through 2. Below 2, roots are complex and the path spirals. At 2, the root repeats and the system returns without oscillating. Above 2, two negative real modes return even more slowly. Critical damping is the boundary between oscillatory and non-oscillatory return.

Worked example: read roots, then constants

Fade the root cases

Match each equation to a root picture and solution form.

Mechanical language

The mass-spring-damper model

$$m{x}^{\prime \prime }+c{x}^{\prime }+kx=0$$

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.

Error clinic

Practice

1. 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

2. Practice - repeated root

   7.2 Solve \(y''+6y'+9y=0\), \(y(0)=0\), \(y'(0)=2\).

   Detailed solution

3. 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

4. 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

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:

$$a{y}^{\prime \prime }+b{y}^{\prime }+cy=g(t).$$

Every solution splits into

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

where (y_h) is transient natural motion and (y_p) is one particular response to the forcing.

Match the input's shape

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

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

Lab: sweep through resonance

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

$$y^{\prime \prime }+y=\cos t,$$

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.

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}).

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

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.

Practice

1. Practice - polynomial input

   8.1 Find the general solution of \(y''+y=t\).

   Detailed solution

2. Practice - sinusoidal input

   8.2 Solve \(y''+4y=\cos t\), \(y(0)=0\), \(y'(0)=0\).

   Detailed solution

3. 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

4. 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

Retrieve the design logic

Explain without formulas:

1. why a particular solution should resemble the input;
2. why a collision with natural motion requires a factor of (t); and
3. 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 rather than one number:

$${\mathbf{x}}^{\prime }=A\mathbf{x},\mathbf{x}=\left[\begin{array}{c}x\\ y\end{array}\right].$$

At every point in the phase plane, the system assigns a velocity vector. A solution is a trajectory tangent to that field.

Lab: change the eigenvalue pattern

Ask three questions for each preset:

1. Do trajectories approach or leave the origin?
2. Do they rotate?
3. 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

$${\mathbf{x}}^{\prime }=\lambda e^{\lambda t}\mathbf{v}=e^{\lambda t}A\mathbf{v}=A\mathbf{x}.$$

Starting on an eigenvector line keeps the trajectory on that line. The eigenvalue says how motion along that direction grows, decays, or oscillates.

Worked example: a saddle

The title deliberately invites a mistake. Classification comes from eigenvalue signs, not from the visual appearance of the matrix.

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}}).

Second-order equations are systems in disguise

Let (x_1=y) and (x_2=y'). Then

$$y^{\prime \prime }+a{y}^{\prime }+by=0$$

becomes

$$\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -b& -a\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right].$$

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.

Error clinic

Practice

1. Practice - diagonal system

   9.1 Solve and sketch \(x'=-x,\ y'=-3y\). Which eigen-direction dominates late?

   Detailed solution

2. 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

3. 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

4. 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

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 elementary closed-form solution. Numerical methods turn the rate rule into a sequence of approximations.

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

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

It is the tangent-line approximation repeated.

Lab: shrink the step

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

A table is part of the reasoning

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 is proportional to (h^2). Over about (1/h) steps across a fixed time interval, these errors accumulate into global error proportional to (h).

Practical consequence: halving (h) should roughly halve Euler's global error once the steps are small enough for the asymptotic pattern to appear.

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 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.

Fade the computation

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

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

Practice

1. 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

2. 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

3. 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

4. Challenge - step stability

   10.4 For \(y'=-ay\) with \(a>0\), find all \(h>0\) for which Euler approximations decay in magnitude. Which subset decays without alternating sign?

   Detailed solution

Retrieve the whole loop

Choose one equation from any earlier chapter and state:

1. its structural fingerprint;
2. a qualitative prediction;
3. an analytic method, if available;
4. a numerical fallback; and
5. 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:

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

Round A: recognize and predict

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

1. Mixed - recognition

   M1 \(y'=ty+\sin t\)

   Detailed solution

2. Mixed - recognition

   M2 \(y'=e^t/(1+y^2)\)

   Detailed solution

3. Mixed - recognition

   M3 \(y''+6y'+9y=t\)

   Detailed solution

4. Mixed - recognition

   M4 \(\mathbf{x}'=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\mathbf{x}\)

   Detailed solution

5. Mixed - qualitative

   M5 \(y'=(y-1)(y+2)\)

   Detailed solution

6. Mixed - verification

   M6 A student claims \(y=1/(1-t)\) solves \(y'=y\), \(y(0)=1\). Diagnose the first mismatch.

   Detailed solution

Round B: execute and verify

1. 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

2. Mixed - first order

   M8 Solve \(y'+(2/t)y=t^2\), \(t>0\), \(y(1)=1\).

   Detailed solution

3. Mixed - second order

   M9 Solve \(y''+2y'+2y=0\), \(y(0)=1\), \(y'(0)=0\). Describe its phase portrait.

   Detailed solution

4. Mixed - forcing

   M10 Find the general solution of \(y''+y=\sin3t\).

   Detailed solution

5. 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

6. 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.

1. Build the differential equation with units.
2. Find and classify the equilibrium concentration.
3. Solve for (C(0)=0).
4. Express the time to reach 90% of equilibrium.
5. 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):

$$P_1^{\prime }=P_1,P_2^{\prime }=P_2(1-P_2/10).$$

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.

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:

1. What do you inspect before choosing a method?
2. Which visual representation would you use for one autonomous state? For two coupled states?
3. How do you distinguish transient and forced behavior?
4. What evidence makes a numerical result credible?
5. 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

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

$$W^{\prime }=12-0.04W.$$

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

$$S^{\prime }=0.24-0.015S,S(0)=10.$$

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

Chapter 2

Solution 2.1

The equilibrium is (y=3). Below 3, (y-3<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)>0), so the solution rises toward the stable equilibrium 1. Differentiate along a solution:

$$y^{\prime \prime }=y^{\prime }(1-2y).$$

While (0<y<1/2), both factors are positive, so the graph is concave up. After (y) passes (1/2), (1-2y<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.

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 simply the shorter option for the first.

Solution 3.3

On (t>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.

Chapter 4

Solution 4.1

Separate and integrate:

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

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

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

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

Solution 4.2

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

$$\int \frac{dy}{y(y-2)}=t+C\Rightarrow \frac{1}{2}\ln \left|\frac{y-2}{y}\right|=t+C.$$

Using (y(0)=1) gives ((y-2)/y=-e^{2t}), 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

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

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

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

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

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

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>0), (\mu=t^2). Then

$$(t^{2}y{)}^{\prime }=t^3.$$

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

$$t=4\ln 6\approx 7.17\text{ time units}.$$

Solution 5.4

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

$$y^{\prime }+y=(2t-C{e}^{-t})+(t^2+C{e}^{-t})=t^2+2t.$$

Thus one valid equation is

$$y^{\prime }+y=t^2+2t.$$

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

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. For the outer initial states, the cubic rate also warns that unbounded growth in magnitude may occur in finite time.

Solution 6.3

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

$$\frac{d{P}^{\prime }}{dP}=r-\frac{2r}{K}P.$$

The critical state is (P=K/2), and the downward-opening parabola makes it a maximum. The maximum rate is (rK/4). The largest sustainable population is the equilibrium (K); these are different quantities.

Solution 6.4

For (a>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<0), (y'=y^2+|a|&gt0), so no equilibria exist and every solution rises.

Chapter 7

Solution 7.1

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

$$y=C_1{e}^{3t}+C_2{e}^{-2t}.$$

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

$$y=(C_1+C_{2}t)e^{-3t}.$$

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).

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

$$y=\frac{\cos t-\cos 2t}{3}.$$

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

Near resonance, the response contains a difference such as (\cos(\omega t)-\cos(\omega_0t)). The identity rewrites it as a fast oscillation with average frequency ((\omega+\omega_0)/2), multiplied by a slow envelope (\sin((\omega-\omega_0)t/2)). When the frequencies are close, the envelope varies slowly, producing beats. As they meet, the limiting envelope grows proportionally to (t).

Chapter 9

Solution 9.1

The system is already diagonal:

$$x=C_1{e}^{-t},y=C_2{e}^{-3t}.$$

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

$$\mathbf{x}=C_1{e}^{3t}\left[\begin{array}{c}1\\ 1\end{array}\right]+C_2{e}^t\left[\begin{array}{c}1\\ -1\end{array}\right].$$

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

$$\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -5& -2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right].$$

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<0), eigenvalues have opposite signs: saddle. If (\Delta>0) and (D>0), the roots are real with common sign set by (\tau): stable node for (\tau<0), unstable node for (\tau>0). If (D<0), they are complex: spiral sink for (\tau<0), spiral source for (\tau>0), and center in the ideal linear case (\tau=0). The curves (\Delta=0) and (D=0) are classification boundaries.

Chapter 10

Solution 10.1

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

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

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<P_n<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

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

Decay without alternating sign requires (0\le1-ah<1), hence

$$0<h\le \frac{1}{a}.$$

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

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:

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

At (t=0,y=-1), (C=1), so

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

The quadratic denominator has discriminant (1-2<0), so it never vanishes; the solution exists for all real (t).

Solution M8

On (t>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

$$y=C_1\cos t+C_2\sin t-\frac{1}{8}\sin 3t.$$

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

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

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_=I/k), stable because the displacement equation is ((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:

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

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

$$A=\left[\begin{array}{cc}-0.5& -1\\ 1& -0.5\end{array}\right].$$

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

$$x^{\prime \prime }+x^{\prime }+1.25x=0.$$

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

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 support early schema acquisition, 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:

1. see a complete example;
2. complete its missing late steps;
3. complete a similar problem with earlier steps removed; and
4. 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 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 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:

1. classify unfamiliar equations without chapter labels;
2. translate among rate laws, fields, and solution behavior;
3. retain methods after a delay;
4. detect realistic erroneous solutions;
5. choose analytic or numerical tools appropriately; and
6. 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.