Three weeks, fifteen focused lessons, and three checkpoint quizzes that take you from "what even is an ODE" to solving systems with eigenvalues and Laplace transforms — so day one of the real course feels like review.
Then a fourth track: the Math 275 course companion, following Prof. Williams' sections §1.1–§2.4 with every classwork problem solved step by step.
Do one lesson per weekday (about an hour). Read the explanation, follow the worked example, then attempt every practice problem before revealing the solution — the struggle is where the learning happens.
Tick the "Mark complete" box at the bottom of each lesson. Your progress bar and checkmarks are saved in this browser automatically.
At the end of each week, take the quiz. Aim for 80%+ before moving on; if you miss one, reread that lesson's worked example.
Keep paper and pencil next to you. ODEs are a doing subject, not a reading subject.
Prerequisites — a quick gut-check
You'll lean hardest on integration techniques (u-substitution, integration by parts, partial fractions), derivatives (product/quotient/chain rule), and a little linear algebra (determinants, eigenvalues) in Week 3. If any of those feel rusty, brush them up alongside Week 1 — nothing here requires them to be perfect on day one.
Rusty on the calculus? The Calculus Toolkit for ODEs covers exactly those prerequisites — differentiation, integration techniques, and series — with interactive diagrams and hint-by-hint problems.
Week 1 · Lesson 1 Foundations
What is an ODE?
Before you solve anything, you need the vocabulary. Half of a differential equations exam is just correctly classifying what's in front of you, because the classification tells you which method to reach for.
①The core idea
A differential equation is just an equation that relates a function to its own derivatives. Instead of solving for a number (like \(x=3\)), you're solving for a whole function \(y(x)\).
An ordinary differential equation (ODE) involves derivatives with respect to a single independent variable. If there were several independent variables and partial derivatives, it would be a partial differential equation (PDE). This course is entirely ODEs.
The order is the highest derivative that appears. \(y' = 3y\) is first-order; \(y'' + 4y = 0\) is second-order.
Linear vs. nonlinear
An ODE is linear if the unknown function \(y\) and all its derivatives appear only to the first power, are not multiplied together, and sit inside no nonlinear function (no \(\sin y\), \(y^2\), \(y\,y'\), etc.). The coefficients may depend on \(x\) freely.
\(y' + x^2 y = \cos x\) → linear (coefficient \(x^2\) is fine — it multiplies \(y\), not a power of \(y\)).
\(y' + y^2 = x\) → nonlinear (the \(y^2\)).
\(y\,y'' = 1\) → nonlinear (product of unknowns).
Homogeneous vs. nonhomogeneous
For a linear ODE, if the "forcing" term \(g(x)=0\) it is homogeneous; otherwise nonhomogeneous. (Heads up: "homogeneous" means something different for first-order substitution problems in Lesson 5 — context tells you which.)
③Solutions, general vs particular
A solution is any function that makes the equation true when substituted in. Because integrating introduces constants, a first-order ODE has a general solution with one arbitrary constant \(C\) — an infinite family of curves. Pin down \(C\) with an initial condition like \(y(0)=2\) and you get a particular solution.
IVP vs BVP
An initial value problem (IVP) gives all conditions at one point (e.g. \(y(0)\) and \(y'(0)\)). A boundary value problem (BVP) gives conditions at different points (e.g. \(y(0)\) and \(y(1)\)). This course is mostly IVPs.
⌖ Worked example
Classify \( \dfrac{d^2y}{dx^2} + 5\dfrac{dy}{dx} + 6y = e^{x} \) and state how many arbitrary constants its general solution has.
STEP 1 — Order
Highest derivative is \(y''\), so it is second-order.
STEP 2 — Linearity
\(y, y', y''\) each appear to the first power, no products, no nonlinear functions of \(y\). The coefficients \(1,5,6\) are constants. So it is linear with constant coefficients.
STEP 3 — Homogeneity
The right side is \(e^x \neq 0\), so it is nonhomogeneous.
STEP 4 — Constants
An \(n\)th-order ODE's general solution carries \(n\) arbitrary constants. Here \(n=2\), so two constants \(C_1, C_2\).
✎Practice
1
Classify by order, linearity, and homogeneity: \( y''' + x\,y' - y = 0 \).
Solution
Highest derivative \(y'''\) → third-order. Each term linear in \(y\) and derivatives, coefficient \(x\) allowed → linear. Right side \(=0\) → homogeneous. Third-order, linear, homogeneous.
2
Linear or nonlinear? (a) \(y' = \sin(x)\,y\) (b) \(y' = \sin(y)\) (c) \( (y')^2 + y = x \).
Solution
(a) Linear — \(\sin x\) is just a coefficient multiplying \(y\). (b) Nonlinear — \(y\) sits inside \(\sin\). (c) Nonlinear — the derivative is squared.
3
Verify that \(y = 3e^{-2x}\) is a solution of \(y' + 2y = 0\).
Solution
\(y' = -6e^{-2x}\). Then \(y' + 2y = -6e^{-2x} + 2(3e^{-2x}) = -6e^{-2x}+6e^{-2x} = 0.\) ✓ It satisfies the equation, so yes.
4
The general solution of \(y' = 6x\) is \(y = 3x^2 + C\). Find the particular solution with \(y(1) = 5\).
Solution
Plug in: \(5 = 3(1)^2 + C \Rightarrow C = 2\). So \(y = 3x^2 + 2\).
⌕References
Boyce & DiPrima, Elementary Differential Equations — §1.1–1.3 (the standard text for this material).
Paul's Online Math Notes — "Differential Equations → Basic Concepts." Excellent free worked examples: tutorial.math.lamar.edu
MIT 18.03 (OCW) — Lecture 1, "The geometrical view of \(y'=f(x,y)\)."
3Blue1Brown — "Differential equations, a tourist's guide" (YouTube) for intuition.
Week 1 · Lesson 2 Foundations
Slope fields, verification & uniqueness
Even when you can't solve an ODE by hand, the equation \(y'=f(x,y)\) tells you the slope of the solution at every point. That geometric picture — plus knowing when a unique solution exists — keeps you honest.
①Verifying a proposed solution
The single most reliable skill in this course: given a candidate \(y(x)\), differentiate it and substitute into both sides. If the equation reduces to a true identity, it's a solution. This is how you check your own work on every problem.
②Slope (direction) fields
At each point \((x,y)\) in the plane, \(f(x,y)\) gives a slope. Draw a short line segment with that slope at a grid of points and you get a slope field. Any solution curve must flow tangent to these segments — so you can sketch solution behavior with zero integration.
Reading a slope field
Follow the little segments like current in a river. An initial condition \((x_0,y_0)\) drops you at one point; the unique solution is the curve threading through the field from there.
③Autonomous equations & equilibria
If \(f\) depends on \(y\) only — \(y' = f(y)\) — the equation is autonomous. Setting \(f(y)=0\) gives equilibrium (constant) solutions. An equilibrium is stable if nearby solutions move toward it, unstable if they run away.
You can't always solve an ODE, but you can often guarantee a unique solution exists. For the IVP \(y' = f(x,y),\; y(x_0)=y_0\):
Existence & Uniqueness (first-order)
If \(f\) and \(\partial f/\partial y\) are both continuous on a rectangle around \((x_0,y_0)\), then a unique solution exists on some interval around \(x_0\).
Why you care
If \(\partial f/\partial y\) blows up, uniqueness can fail. Classic case: \(y' = y^{1/3},\; y(0)=0\) has multiple solutions through the origin. Spotting this prevents you from trusting a single answer that isn't the whole story.
⌖ Worked example
Does \(y' = \dfrac{x}{y},\; y(0)=2\) have a unique solution near \(x=0\)? Then verify \(y=\sqrt{x^2+4}\) solves it.
STEP 1 — Check the hypotheses
\(f(x,y)=x/y\) and \(\partial f/\partial y = -x/y^2\) are both continuous as long as \(y\neq 0\). At \((0,2)\), \(y=2\neq0\), so both are continuous nearby → unique solution exists.
Verify that \(y = C e^{x} - x - 1\) solves \(y' = x + y\) for any constant \(C\).
Solution
\(y' = Ce^{x} - 1\). And \(x + y = x + Ce^x - x - 1 = Ce^x - 1\). Both sides equal \(Ce^x-1\). ✓
2
Find all equilibrium solutions of \(y' = y^2 - 4\) and classify their stability.
Solution
\(y^2-4=0\Rightarrow y=\pm2\). For \(y>2\), \(y'>0\) (grows); between \(-2\) and \(2\), \(y'<0\) (falls); below \(-2\), \(y'>0\). So arrows point away from \(y=2\) (unstable) and toward \(y=-2\) (stable).
3
Where might the IVP \(y' = \sqrt{y},\; y(0)=0\) fail to have a unique solution, and why?
Solution
\(f=\sqrt y\), \(\partial f/\partial y = \tfrac{1}{2\sqrt y}\), which is not continuous at \(y=0\). The uniqueness hypothesis fails there, and indeed both \(y\equiv0\) and \(y=\tfrac14 x^2\) (for \(x\ge0\)) satisfy the IVP. Uniqueness is not guaranteed.
4
A slope field for \(y'=-y\) shows segments that are flat on the \(x\)-axis and steepen away from it. Sketch the behavior of the solution with \(y(0)=3\).
Solution
Starting at \((0,3)\), slopes are negative above the axis, so the curve decays toward \(y=0\) as \(x\) increases (and grows for \(x<0\)). It's exponential decay \(y=3e^{-x}\) — \(y=0\) is a stable equilibrium the curve approaches but never crosses.
Paul's Online Math Notes — "Direction Fields" and "Equilibrium Solutions."
Desmos / Slopefield plotters — type \(y'=f(x,y)\) into an online slope-field grapher and experiment; it builds intuition fast.
Week 1 · Lesson 3 First-order methods
Separable equations
The first method you can actually solve with. If you can algebraically peel the \(x\)'s onto one side and the \(y\)'s onto the other, you just integrate both sides — done.
①The shape to look for
An ODE is separable if it can be written as a product of a function of \(x\) and a function of \(y\):
Separable form
\[ \frac{dy}{dx} = g(x)\,h(y) \]
Treat \(dy/dx\) as a ratio of differentials (legitimate here as shorthand for the chain rule), gather like with like, and integrate:
Write \(y'\) as \(dy/dx\) and get all \(y\)-stuff (with \(dy\)) on one side, all \(x\)-stuff (with \(dx\)) on the other.
Integrate both sides. Add a single constant \(C\) (on the \(x\) side, by convention).
Apply the initial condition now if you have one — solve for \(C\).
Solve for \(y\) explicitly if possible (sometimes you must leave it implicit).
Two traps
(a) Dividing by \(h(y)\) can kill constant solutions where \(h(y)=0\). Check those separately — they're often legitimate equilibrium solutions. (b) Don't forget the \(+C\); it's where the whole solution family lives.
Radioactive decay: \(\dfrac{dN}{dt} = -kN\) with \(N(0)=N_0\). Solve, and find the half-life.
Solution
Separable: \(\dfrac{dN}{N}=-k\,dt \Rightarrow \ln N = -kt + C \Rightarrow N = N_0 e^{-kt}.\) Half-life when \(N=\tfrac12 N_0\): \( \tfrac12 = e^{-kt_{1/2}} \Rightarrow t_{1/2} = \dfrac{\ln 2}{k}.\)
⌕References
Boyce & DiPrima — §2.2 Separable Equations.
Paul's Online Math Notes — "Separable Equations" (lots of fully worked problems).
Khan Academy — "Separable equations" unit, video + practice set.
Week 1 · Lesson 4 First-order methods
Linear equations & integrating factors
The most important first-order technique you'll use all term. Any first-order linear ODE — even ones that aren't separable — yields to one slick trick: multiply by an integrating factor so the left side collapses into a single derivative.
①Standard form first
Always rewrite the equation as:
Standard form
\[ \frac{dy}{dx} + P(x)\,y = Q(x) \]
The leading coefficient on \(y'\) must be \(1\). Divide through if it isn't. Getting this wrong is the #1 source of errors here.
②The integrating factor
Multiply everything by \(\mu(x)\), chosen so the left side becomes the derivative of a product:
Mixing tank: a tank holds 100 L of pure water; brine with 2 g/L of salt flows in at 5 L/min and the well-mixed solution flows out at 5 L/min. Let \(S(t)\) be grams of salt. Set up and solve.
Solution
Rate in \(=2\cdot5=10\) g/min; rate out \(=\dfrac{S}{100}\cdot5=\dfrac{S}{20}.\) So \(S' + \tfrac{1}{20}S = 10,\; S(0)=0.\) \(\mu=e^{t/20}\): \(\dfrac{d}{dt}[e^{t/20}S]=10e^{t/20}\Rightarrow e^{t/20}S=200e^{t/20}+C.\) So \(S=200+Ce^{-t/20}\); IC gives \(C=-200\). \(\;S(t)=200\big(1-e^{-t/20}\big),\) approaching 200 g.
⌕References
Boyce & DiPrima — §2.1 Linear Equations; Integrating Factors. §2.3 for the modeling/mixing applications.
Paul's Online Math Notes — "Linear Differential Equations" (the integrating-factor walkthrough is gold).
MIT 18.03 — Lecture on first-order linear ODEs and the "input–response" view.
Week 1 · Lesson 5 First-order methods
Exact equations & substitutions
Two more arrows for your first-order quiver: exact equations (recognizing a hidden total derivative) and substitutions that convert a scary nonlinear equation into one you already know how to solve. Then it's quiz time.
①Exact equations
Write the ODE as \(M(x,y)\,dx + N(x,y)\,dy = 0\). It is exact if it's secretly \(d[F(x,y)] = 0\) for some potential function \(F\). The test:
Paul's Online Math Notes — "Exact Equations" and "Bernoulli Differential Equations."
Trefor Bazett (YouTube) — concise exact-equation and substitution walkthroughs.
Next up
You've now got every standard first-order method. Take Quiz 1 to lock it in before moving to second-order equations.
Week 1 · Checkpoint Quiz
Quiz 1 — First-Order ODEs
Six questions on classification and the first-order toolkit. Tap an answer to lock it in and see why. Aim for 5/6 before starting Week 2.
Question 1
Classify \( y'' + (\sin x)\,y = x^3 \).
Why
Highest derivative is \(y''\) (second-order). \(\sin x\) is just a coefficient on \(y\), so it stays linear. The right side \(x^3\neq0\) makes it nonhomogeneous.
Question 2
Which method fits \( \dfrac{dy}{dx} = \dfrac{x^2}{1+y^2} \) most directly?
Why
It separates cleanly: \((1+y^2)\,dy = x^2\,dx\). Integrate to \(y + \tfrac13 y^3 = \tfrac13 x^3 + C\).
Question 3
For \( y' + 2y = x \), the integrating factor is:
Why
\(P(x)=2\), so \(\mu=e^{\int 2\,dx}=e^{2x}\).
Question 4
The solution of \( y' = ky,\; y(0)=y_0 \) is:
Why
The canonical exponential growth/decay solution. Separate: \(dy/y = k\,dt \Rightarrow \ln y = kt + C \Rightarrow y=y_0 e^{kt}\).
Question 5
\( (2x + y)\,dx + (x + 2y)\,dy = 0 \) — is it exact?
Why
\(M=2x+y\Rightarrow M_y=1\); \(N=x+2y\Rightarrow N_x=1\). Equal, so exact. (Solution: \(x^2+xy+y^2=C\).)
Question 6
For the autonomous equation \( y' = y(y-3) \), the equilibrium \(y=0\) is:
Why
Near \(y=0\): for small \(y>0\), \(y'=y(y-3)<0\) (falls back); for \(y<0\), \(y'>0\) (rises toward 0). Arrows point inward → stable. (And \(y=3\) is unstable.)
Week 2 · Lesson 6 Second-order theory
Linear theory: superposition & the Wronskian
Second-order linear ODEs run most of physics and engineering — vibrations, circuits, control systems. Before grinding out solutions, you need the structural rules that tell you when you have the complete solution.
①The structure of the solution
For a linear second-order ODE \( y'' + p(x)y' + q(x)y = g(x) \), the general solution always splits into two pieces:
where \(y_1, y_2\) are two independent solutions of the homogeneous equation (\(g=0\)) and \(y_p\) is any single solution of the full equation. Weeks 2 is largely about finding each piece.
②Superposition (homogeneous only)
If \(y_1\) and \(y_2\) both solve the homogeneous equation, so does any combination \(C_1 y_1 + C_2 y_2\). This is the superposition principle — and it's exactly why linearity is so powerful.
③Linear independence & the Wronskian
Two functions are linearly independent if neither is a constant multiple of the other. To get the general solution you need two independent ones. The Wronskian tests this:
If \(W \neq 0\) at some point in the interval, \(y_1\) and \(y_2\) are linearly independent and form a fundamental set — their combination is the full homogeneous solution.
⌖ Worked example
Show \(y_1=e^{2x}\) and \(y_2=e^{-x}\) are independent solutions of \(y'' - y' - 2y = 0\), and write the general solution.
Compute the Wronskian of \(y_1=\cos x,\;y_2=\sin x\). Are they independent?
Solution
\(W = \cos x\cos x - \sin x(-\sin x) = \cos^2 x + \sin^2 x = 1 \neq 0.\) Independent.
2
Are \(y_1=e^{x}\) and \(y_2=3e^{x}\) linearly independent?
Solution
No — \(y_2=3y_1\) is a constant multiple. Check: \(W=e^x(3e^x)-3e^x(e^x)=0\). Dependent; they do not form a fundamental set.
3
If \(y_p = x\) solves \(y'' + y = x\) and the homogeneous solution is \(C_1\cos x + C_2\sin x\), write the general solution.
Solution
\(y = C_1\cos x + C_2\sin x + x.\) (Check: \(y_p''=0\), \(y_p''+y_p = 0 + x = x\) ✓.)
4
How many initial conditions do you need to pin down a unique solution of a second-order linear IVP, and what's the typical form?
Solution
Two — because there are two constants \(C_1,C_2\). Typically \(y(x_0)=a\) and \(y'(x_0)=b\) (position and velocity, physically).
⌕References
Boyce & DiPrima — §3.2 Fundamental Solutions & the Wronskian; §3.1 overview.
Paul's Online Math Notes — "Fundamental Sets of Solutions" and "More on the Wronskian."
MIT 18.03 — Lectures on second-order linear ODEs and the structure theorem.
Week 2 · Lesson 7 Second-order
Constant coefficients: real roots
The workhorse calculation of the whole course. For \(ay'' + by' + cy = 0\), guessing \(y=e^{rx}\) turns calculus into algebra — you just solve a quadratic.
①The characteristic equation
Substitute \(y=e^{rx}\) (so \(y'=re^{rx}, y''=r^2 e^{rx}\)) into \(ay''+by'+cy=0\) and divide by \(e^{rx}\):
Characteristic (auxiliary) equation
\[ a r^2 + b r + c = 0 \]
The roots \(r\) determine everything. There are three cases by the discriminant \(b^2-4ac\). This lesson handles the two real cases; complex roots are Lesson 8.
②Case 1 — two distinct real roots
If \(b^2 - 4ac > 0\), you get \(r_1 \neq r_2\), both real:
\[ y = C_1 e^{r_1 x} + C_2 e^{r_2 x} \]
③Case 2 — one repeated real root
If \(b^2-4ac = 0\), you get a single root \(r\) (twice). One solution is \(e^{rx}\); the second independent solution gains a factor of \(x\):
\[ y = C_1 e^{r x} + C_2\, x\, e^{r x} \]
Don't lose the second solution
With a repeated root you must include the \(xe^{rx}\) term, or you'll only have one constant and can't satisfy two initial conditions.
\(4r^2+4r+1=(2r+1)^2=0\Rightarrow r=-\tfrac12\) double. \(y=C_1 e^{-x/2}+C_2 x e^{-x/2}.\)
⌕References
Boyce & DiPrima — §3.1 Homogeneous Equations with Constant Coefficients; §3.4 Repeated Roots; Reduction of Order.
Paul's Online Math Notes — "Real, Distinct Roots" and "Repeated Roots."
Week 2 · Lesson 8 Second-order
Complex roots & oscillation
When the characteristic quadratic has complex roots, the solutions become sines and cosines — this is where ODEs start to vibrate. Euler's formula is the bridge.
①Case 3 — complex conjugate roots
If \(b^2 - 4ac < 0\), the roots come as a conjugate pair \(r = \alpha \pm \beta i\). Using Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\), the complex exponentials recombine into real oscillations:
Complex-root solution
\[ r = \alpha \pm \beta i \;\Longrightarrow\; y = e^{\alpha x}\big( C_1\cos\beta x + C_2 \sin\beta x \big) \]
The real part \(\alpha\) controls growth/decay of the envelope; the imaginary part \(\beta\) sets the oscillation frequency.
Physical reading
\(\alpha<0\): a damped, decaying oscillation (a real spring losing energy). \(\alpha=0\): pure undamped oscillation. \(\alpha>0\): a growing oscillation (instability).
Without solving, predict the long-term behavior of solutions to \( y'' - 2y' + 10y = 0 \).
Solution
\(r=\dfrac{2\pm\sqrt{4-40}}{2}=1\pm3i\). Real part \(\alpha=+1>0\), so solutions oscillate with a growing envelope \(e^{x}\) — they blow up.
⌕References
Boyce & DiPrima — §3.3 Complex Roots of the Characteristic Equation.
Paul's Online Math Notes — "Complex Roots."
3Blue1Brown — "What is Euler's formula actually saying?" for the \(e^{i\theta}\) intuition.
Week 2 · Lesson 9 Nonhomogeneous
Undetermined coefficients
Now the right side isn't zero. To find a particular solution \(y_p\) you make an educated guess that mirrors the forcing term \(g(x)\), then solve for the coefficients. Fast and clean — when \(g\) is "nice."
①The strategy
For \(ay'' + by' + cy = g(x)\): the full solution is \(y = y_h + y_p\). You already know how to get \(y_h\) (Lessons 7–8). For \(y_p\), guess a form that looks like \(g(x)\) with unknown coefficients, plug in, and match.
Forcing term \(g(x)\)
Guess for \(y_p\)
\(e^{kx}\)
\(A e^{kx}\)
polynomial deg \(n\)
\(A_n x^n + \cdots + A_1 x + A_0\)
\(\sin kx\) or \(\cos kx\)
\(A\cos kx + B\sin kx\)
\(e^{kx}\sin kx\), etc.
product of the above
②The overlap rule (crucial)
If your guess already appears in \(y_h\), multiply by \(x\)
If a term in your trial \(y_p\) duplicates a homogeneous solution, multiply that trial term by \(x\) (or \(x^2\) for a double root) until it no longer overlaps. Forgetting this is the single most common mistake on this topic.
\(r=(r+1)^2=0\Rightarrow y_h=C_1 e^{-x}+C_2 x e^{-x}.\) Both \(e^{-x}\) and \(xe^{-x}\) overlap, so try \(y_p=Ax^2 e^{-x}.\) Substituting gives \(2A e^{-x}=2e^{-x}\Rightarrow A=1.\) \(y=C_1 e^{-x}+C_2 x e^{-x}+x^2 e^{-x}.\)
Paul's Online Math Notes — "Undetermined Coefficients" (the trial-form table and overlap rule).
Week 2 · Lesson 10 Nonhomogeneous · Applications
Variation of parameters & vibrations
Undetermined coefficients only works for "nice" forcing terms. Variation of parameters is the universal method — it works for any \(g(x)\). Then we cash everything in on the spring–mass system that models half of engineering.
①Variation of parameters
Given \(y_h = C_1 y_1 + C_2 y_2\), look for a particular solution \(y_p = u_1 y_1 + u_2 y_2\) where \(u_1, u_2\) are now functions. Imposing a convenient condition leads to:
Variation of parameters (standard form \(y''+p y'+q y=g\))
\(y_p = \cos x\ln|\cos x| + x\sin x.\) So \(\boxed{y = C_1\cos x + C_2\sin x + \cos x\ln|\cos x| + x\sin x}.\)
②Application: mechanical vibrations
A mass \(m\) on a spring (constant \(k\)) with damping \(c\) and external force \(F(t)\) obeys:
The spring–mass–damper equation
\[ m\,x'' + c\,x' + k\,x = F(t) \]
Undamped, free (\(c=0,F=0\)): \(x''+\omega_0^2 x=0\), pure oscillation at \(\omega_0=\sqrt{k/m}\).
Damped: the discriminant of the characteristic equation sorts it into underdamped (oscillates & decays — complex roots), critically damped (double root), or overdamped (two real negative roots, no oscillation).
Forced: a periodic \(F(t)\) drives the system; matching the natural frequency gives resonance (the \(x\cdot\) overlap term from Lesson 9).
Same math, many systems
The identical equation models an RLC circuit: \(L q'' + R q' + \tfrac1C q = E(t)\). Mastering the spring gets you the circuit for free.
⌖ Worked example (vibration)
A 1 kg mass on a spring with \(k=4\) N/m and damping \(c=4\) N·s/m, no forcing. Classify and solve \(x'' + 4x' + 4x = 0\).
Classify the damping for \( x'' + 5x' + 6x = 0 \).
Solution
\(r^2+5r+6=(r+2)(r+3)=0\Rightarrow r=-2,-3\): two distinct real negative roots → overdamped. \(x=C_1 e^{-2t}+C_2 e^{-3t}.\)
3
Find the natural frequency of \( 2x'' + 18x = 0 \).
Solution
Divide by 2: \(x''+9x=0\Rightarrow \omega_0^2=9\Rightarrow \omega_0=3\) rad/s. \(x=C_1\cos 3t+C_2\sin 3t.\)
4
Which forcing frequency \(\omega\) drives resonance in \( x'' + 16x = \cos\omega t \)?
Solution
Natural frequency \(\omega_0=\sqrt{16}=4\). Resonance when \(\omega=\omega_0=4\): the trial \(\cos 4t\) overlaps \(y_h\), forcing a \(t\)-growing solution.
Paul's Online Math Notes — "Variation of Parameters" and "Mechanical Vibrations."
MIT 18.03 — Lectures on damping, resonance, and the RLC analogy.
Next up
You've conquered second-order linear ODEs. Quiz 2 awaits before the Week 3 power tools.
Week 2 · Checkpoint Quiz
Quiz 2 — Second-Order Linear ODEs
Six questions on characteristic roots, undetermined coefficients, and vibrations. Tap to answer and reveal the reasoning.
Question 1
The general solution of \( y'' - 7y' + 12y = 0 \) is:
Why
\(r^2-7r+12=(r-3)(r-4)=0\Rightarrow r=3,4\): distinct real roots.
Question 2
Roots \(r = -1 \pm 2i\) give a general solution of the form:
Why
\(\alpha=-1,\beta=2\): envelope \(e^{-x}\) times \(\cos/\sin 2x\).
Question 3
A double root \(r=5\) gives:
Why
Repeated root → second solution gains a factor of \(x\).
Question 4
For \( y'' - y = e^{x} \), the correct trial particular solution is:
Why
\(y_h=C_1 e^x+C_2 e^{-x}\); \(e^x\) overlaps, so multiply by \(x\) once (single root).
Question 5
A spring–mass system \( x'' + 6x' + 9x = 0 \) is:
Why
\(r^2+6r+9=(r+3)^2\): a repeated real root is exactly the critically damped case.
Question 6
Which method works for any continuous forcing \(g(x)\), including \(\sec x\) or \(\ln x\)?
Why
Undetermined coefficients needs a "nice" \(g\) (exponentials, polynomials, sin/cos). Variation of parameters has no such restriction.
Week 3 · Lesson 11 Higher order
Higher-order linear equations
Everything from Week 2 generalizes cleanly. A third- or fourth-order constant-coefficient equation just gives a higher-degree characteristic polynomial — the same three root cases, now possibly mixed in one problem.
①The characteristic polynomial
For \(a_n y^{(n)} + \cdots + a_1 y' + a_0 y = 0\), substitute \(y=e^{rx}\) to get a degree-\(n\) polynomial:
How many initial conditions does a unique solution of \( y^{(4)} + \cdots = g(x) \) require?
Solution
Four — equal to the order. Typically \(y(x_0),y'(x_0),y''(x_0),y'''(x_0)\).
⌕References
Boyce & DiPrima — §4.1–4.2 Higher-Order Linear Equations; Homogeneous with Constant Coefficients.
Paul's Online Math Notes — "Higher Order Differential Equations → Basic Concepts."
Week 3 · Lesson 12 Laplace transforms
Laplace transforms: the dictionary
The Laplace transform converts a differential equation in \(t\) into an algebraic equation in \(s\). You'll especially love it for discontinuous or impulsive forcing — exactly the inputs that wreck the earlier methods.
The derivative rule swallows the initial conditions \(f(0), f'(0)\) automatically — so Laplace solves IVPs in one shot, no separate "apply the ICs" step.
Use the \(e^{at}\cos bt\) row with \(a=2,b=1\): \(\dfrac{s-2}{(s-2)^2+1}.\)
4
Using the derivative rule, write \(\mathcal{L}\{y''\}\) for an IVP with \(y(0)=1, y'(0)=-2\), letting \(Y=\mathcal{L}\{y\}\).
Solution
\(\mathcal{L}\{y''\}=s^2 Y - s\,y(0) - y'(0) = s^2 Y - s + 2.\)
⌕References
Boyce & DiPrima — §6.1–6.2 Definition of the Laplace Transform; Solution of Initial Value Problems.
Paul's Online Math Notes — "Laplace Transforms → The Definition" and the full transform table.
Keep a one-page Laplace table handy — most exams provide one, but knowing the top rows cold saves time.
Week 3 · Lesson 13 Laplace transforms
Inverse Laplace & solving IVPs
Forward transform, do algebra in \(s\), then come back with the inverse transform — usually via partial fractions to match the table. This is the complete Laplace workflow for solving an initial value problem.
①The four-step workflow
Transform the whole ODE; use the derivative rule to fold in the initial conditions.
Solve algebraically for \(Y(s)\) — it's just algebra now.
Partial fractions to break \(Y(s)\) into table-matching pieces.
Solve \( y'' + y = 0,\; y(0)=0,\; y'(0)=2 \) using Laplace.
Solution
\(s^2 Y - 0 - 2 + Y = 0\Rightarrow Y=\dfrac{2}{s^2+1}\Rightarrow y=2\sin t.\)
⌕References
Boyce & DiPrima — §6.2 Solving IVPs; review partial fractions from precalculus if rusty.
Paul's Online Math Notes — "Inverse Laplace Transforms" and "Solving IVPs with Laplace Transforms."
Week 3 · Lesson 14 Laplace transforms
Step functions & impulses
The real reason Laplace earns its keep: it handles forcing that switches on suddenly (a unit step) or hammers instantaneously (an impulse) — inputs that break undetermined coefficients but model real switches, jolts, and shocks.
①The unit step (Heaviside) function
Unit step at \(t=a\)
\[ u_a(t) = \begin{cases} 0, & t < a \\ 1, & t \ge a \end{cases} \qquad \mathcal{L}\{u_a(t)\} = \frac{e^{-as}}{s} \]
It lets you write piecewise forcing as one formula — e.g. a force that turns on at \(t=2\) is \(u_2(t)\cdot(\text{something})\).
②The second shifting theorem
Shifting in \(t\) corresponds to multiplying by \(e^{-as}\) in \(s\):
The theorem needs \(f(t-a)\), i.e. the function shifted to start at \(a\). If your term is \(u_a(t)\,g(t)\) with \(g\) not already shifted, rewrite \(g(t)\) in terms of \((t-a)\) first.
③The Dirac delta (impulse)
The Dirac delta \(\delta(t-a)\) models an idealized instantaneous unit impulse at \(t=a\) (a bat hitting a ball, a voltage spike):
Paul's Online Math Notes — "Step Functions," "Solving IVPs with Step Functions," and "Dirac Delta Function."
Week 3 · Lesson 15 Systems · Finale
Systems of first-order ODEs
The grand finale, and where your linear algebra pays off. Coupled equations — predator/prey, multi-tank mixing, coupled circuits — become a single matrix equation \(\mathbf{x}' = A\mathbf{x}\), solved with eigenvalues and eigenvectors.
①Matrix form
A linear system like \(x_1' = a x_1 + b x_2,\; x_2' = c x_1 + d x_2\) packs into:
Linear system
\[ \mathbf{x}' = A\mathbf{x}, \qquad A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
Guided by the scalar case \(x'=ax\Rightarrow x=e^{at}\), we guess vector solutions \(\mathbf{x} = \mathbf{v}\,e^{\lambda t}\). Substituting gives the eigenvalue problem.
So \(\lambda\) is an eigenvalue of \(A\) and \(\mathbf{v}\) its eigenvector. Find eigenvalues from \(\det(A-\lambda I)=0\), then each eigenpair gives one solution. For distinct real eigenvalues:
Find the eigenvalues of \( A = \begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix} \) and write the solution of \(\mathbf x'=A\mathbf x\).
Solution
Diagonal → \(\lambda=3,-2\) with eigenvectors \(\binom10,\binom01\). \(\mathbf x=C_1\binom10 e^{3t}+C_2\binom01 e^{-2t}.\) (The equations are already uncoupled.)
2
For \( A=\begin{pmatrix}0&1\\-2&-3\end{pmatrix} \), find the eigenvalues.
Solution
\(\det\begin{pmatrix}-\lambda&1\\-2&-3-\lambda\end{pmatrix}=\lambda(\lambda+3)+2=\lambda^2+3\lambda+2=(\lambda+1)(\lambda+2)=0\Rightarrow \lambda=-1,-2.\) Both negative → stable node.
3
Convert the second-order ODE \( y'' + 5y' + 6y = 0 \) into a first-order system.
Solution
Let \(x_1=y,\;x_2=y'.\) Then \(x_1'=x_2,\;x_2'=y''=-6y-5y'=-6x_1-5x_2.\) So \(\mathbf x'=\begin{pmatrix}0&1\\-6&-5\end{pmatrix}\mathbf x.\) (Note its eigenvalues \(-2,-3\) match the characteristic roots — same problem, two views.)
4
If a system has eigenvalues \(\lambda = -1 \pm 2i\), describe the phase-plane behavior.
Solution
Complex with negative real part → trajectories spiral inward toward the origin (a stable spiral / spiral sink). The \(\pm2i\) sets the rotation rate.
⌕References
Boyce & DiPrima — §7.1–7.6 Systems of First-Order Linear Equations; the eigenvalue method & phase portraits.
Paul's Online Math Notes — "Systems of Differential Equations → Real Eigenvalues" (and complex/repeated).
3Blue1Brown — "Essence of Linear Algebra" (eigenvectors episode) if eigenvalues feel shaky.
MIT 18.03 — Lectures on systems, matrices, and phase planes.
You made it
That's the full arc of a first ODE course. Take the comprehensive final quiz — if you can clear it, you'll walk into day one already fluent.
Week 3 · Final checkpoint Comprehensive quiz
Quiz 3 — Comprehensive Final
Seven questions spanning all three weeks. This is your readiness check — clear 6/7 and you're ahead of the class.
Question 1 · Methods
Which technique solves \( y' + \dfrac{2}{x}y = x^3 \) most directly?
Why
It's already first-order linear in standard form with \(P=2/x\). Integrating factor \(\mu=x^2\).
Question 2 · Second-order
The solution of \( y'' + 25y = 0 \) is:
Why
\(r^2+25=0\Rightarrow r=\pm5i\): pure oscillation at frequency 5.
Question 3 · Undetermined coefficients
For \( y'' + 4y = \cos 2x \), the correct trial \(y_p\) is:
Why
\(y_h=C_1\cos 2x+C_2\sin 2x\) already contains \(\cos 2x\) — resonance overlap, so multiply by \(x\).
Question 4 · Laplace
\( \mathcal{L}\{y'\} \) for an IVP with \(y(0)=5\) equals:
Why
Derivative rule: \(\mathcal{L}\{y'\}=sY - y(0) = sY - 5\).
Question 5 · Inverse Laplace
\( \mathcal{L}^{-1}\!\left\{\dfrac{3}{s-4}\right\} \) is:
Why
\(\mathcal{L}^{-1}\{1/(s-a)\}=e^{at}\) with \(a=4\), scaled by 3.
Question 6 · Systems
The eigenvalues of \( A=\begin{pmatrix}2&0\\0&5\end{pmatrix} \) are:
Why
A diagonal matrix has its diagonal entries as eigenvalues.
Question 7 · Stability
A system with eigenvalues \( \lambda = -2 \pm 3i \) produces trajectories that:
Why
Complex eigenvalues → spiral; negative real part \(-2\) → inward (stable spiral).
Math 275 · §1.1 Prof. Williams
Definitions & Terminology
The whole first section is vocabulary — but it is the vocabulary that decides which method you reach for later. Every classification question on the exam is answered from this page.
What this track is
These seven lessons follow Prof. Williams' Math 275 section by section: the definitions and worked examples from the course write-ups and lecture notes, then every classwork problem from that section solved step by step.
①What you're assumed to already have
The course opens with an explicit prerequisite list. Nothing here is taught again, so a gap is a gap:
Differentiation
Rates of change · position, velocity, acceleration · power and quotient rules · derivatives of trig functions · chain rule · implicit differentiation · partial derivatives · differentials \((dx,dy,dz)\) · all four notations: Leibniz \(\frac{d}{dx}f(x)\), prime \(f'(x)\), dot \(\dot f(x)\), operator \(D(f(x))\).
Integration
Area under a curve · the inverse operation of differentiation · definite vs. indefinite · \(u\)-substitution · integration by parts · improper integrals · trig substitution · partial fractions.
Series
Convergence and divergence tests · interval of convergence · arithmetic of series · re-indexing a series · power series expansion · differentiating and integrating a series term by term.
A differential equation (DE) is an equation that contains derivatives. Its solution is a function — or a family of functions — \(y\) that satisfies the equation.
The first split is by how many independent variables are involved: an ordinary differential equation (ODE) has one, a partial differential equation (PDE) has several and uses partial derivatives.
③The four classifications
Order
Like the degree of a polynomial: the order is the highest derivative present.
A DE is linear if every function of \(y\) that appears is of degree 1. The coefficients may involve \(x\) however they like — it is the dependent variable that is constrained.
Non-linear examples
\[ yy' - 3xy = x \quad\text{— } yy' \text{ is the non-linear term} \]\[ \frac{d^2y}{dx^2} - y^2 = 3 \quad\text{— } y^2 \text{ is the non-linear term} \]
The one that catches people
\(e^{x}y'\) is linear — \(e^x\) is just a coefficient. \(e^{y}y'\) is non-linear — the dependent variable is inside a function. One letter apart, opposite answers. Classwork problems 1 and 2 are built on exactly this pair.
Homogeneity
Put the equation in standard form (\(y\) and all its derivatives on one side). If the other side is \(0\) it is homogeneous; otherwise non-homogeneous.
An initial value problem (IVP) places its conditions on the initial values of the solution — all at the same point. A boundary value problem (BVP) places conditions at the end boundary, or at both ends.
To check that a function solves a DE you plug it in. Generally solutions come in families — infinitely many functions of the same shape. An initial condition narrows a family to one member; a boundary condition need not.
⌖ Worked example — from the write-up
Is \(y = e^{3x}\cos 2x\) a solution of \(y'' - 6y' + 13y = 0\)?
Both brackets are \(0\), so the whole expression is \(0\). ✓ It is a solution.
STEP 5 — The family version
Repeating the computation with \(y = c_1e^{3x}\cos 2x\) gives \(c_1e^{3x}[(5-18+13)\cos 2x + 0] = 0\) as well. So the family solves it, varying by amplitude; the single function \(y=e^{3x}\cos2x\) is the member picked out by \(y(0)=1,\,y'(0)=3\).
Interval of existence
The interval of existence of a solution is the domain over which the solution actually satisfies the conditions of the DE. If the solution is \(y=\frac1x\) and the condition is \(y(1)=1\), the interval of existence is \(x\in(0,\infty)\) — the branch containing the initial point, not the whole domain of the formula.
✎§1.1 Classwork — all six, worked
1
Consider \( x^2y'' - e^{x}y' + 3y = \sin\!\big(\tfrac12 x\big) + e^{3x} \). (a) Linear or non-linear? (b) Homogeneous or non-homogeneous? (c) What is the order?
Solution
STEP 1 — Find the dependent variable's appearances
They are \(y''\), \(y'\), and \(y\) — each to the first power, none multiplied by another, none inside a function.
STEP 2 — Check the coefficients
\(x^2\), \(-e^{x}\), and \(3\). Every one is a function of the independent variable only, which linearity permits without restriction. The \(e^{x}\) is a coefficient, not a nonlinearity.
STEP 3 — (a)
Linear.
STEP 4 — (b)
The equation is already in standard form and the right side is \(\sin(\tfrac12x) + e^{3x} \neq 0\). Non-homogeneous. (The forcing term involves only \(x\), which is exactly what a forcing term is.)
STEP 5 — (c)
Highest derivative present is \(y''\). Order 2.
Answer
2nd order, linear, non-homogeneous ODE.
2
Consider \( x^2y''' - e^{y}y' + 3y = 0 \). (a) Linear or non-linear? (b) Homogeneous or non-homogeneous? (c) What is the order?
Solution
STEP 1 — Compare with problem 1
Only one character changed: \(e^{x}\) became \(e^{y}\). That single change moves the dependent variable inside a function.
STEP 2 — (a)
\(e^{y}\) is not of degree 1 in \(y\) — its power series is \(1+y+\tfrac{y^2}{2}+\cdots\). And it multiplies \(y'\), so you also have a product of unknowns. Non-linear, with \(e^{y}y'\) as the offending term.
STEP 3 — (b)
Right side is \(0\). Homogeneous.
STEP 4 — (c)
Highest derivative is \(y'''\). Order 3.
STEP 5 — Say it in full
3rd order, homogeneous, non-linear ODE. Note that "homogeneous" is decided independently of "linear" — an equation can be one without the other.
3
Is \( y = c_1e^{3x} + c_2e^{-2x} \) a solution of the ODE \( y'' + y' - 6y = 0 \)?
\(= 6c_1e^{3x} - 4c_2e^{-2x}\), which is not identically zero unless \(c_1=c_2=0\).
STEP 4 — Verdict
No. Only the trivial solution \(y\equiv0\) survives, and that isn't what was asked about.
STEP 5 — What would have worked
Substituting \(y=e^{rx}\) gives \(r^2+r-6=(r+3)(r-2)=0\), so the exponents that do solve it are \(-3\) and \(2\): \(y=c_1e^{-3x}+c_2e^{2x}\). The problem hands you the exponents with their signs swapped. §1.2 problem 1 uses the correct pair — compare them side by side.
4
Is \( y = ce^{2x}\sin^2(x) \) a solution of \( y' - xy = a\cos(x) \)?
Solution
STEP 1 — Differentiate (product + chain rule)
\(y' = 2ce^{2x}\sin^2 x + ce^{2x}\cdot 2\sin x\cos x = 2ce^{2x}\sin x\,(\sin x + \cos x)\).
STEP 2 — Form the left side
\(y' - xy = ce^{2x}\sin x\big[\,2\sin x + 2\cos x - x\sin x\,\big]\).
STEP 3 — Test at a convenient point
At \(x=0\): \(\sin 0 = 0\), so the left side is \(0\); the right side is \(a\cos 0 = a\). Matching forces \(a=0\).
STEP 4 — Now force the whole identity
With \(a=0\) the requirement becomes \(ce^{2x}\sin x\,[2\sin x + 2\cos x - x\sin x] \equiv 0\). The bracket is not identically zero (at \(x=\pi/2\) it equals \(2-\pi/2 \approx 0.43\)), and \(e^{2x}\) never vanishes, so the only escape is \(c=0\).
STEP 5 — Verdict
No — not for any non-trivial \(c\), and not for any constant \(a\). The point of the exercise is the procedure: differentiate, substitute, and compare as an identity in \(x\), not at a single point.
Note
The right side of this one as printed on the sheet mixes an unexplained constant \(a\) into an otherwise \(y\)-and-\(x\) equation. Whichever way it was intended, the verification method above is the graded skill — and the honest answer to the printed equation is "no."
5
Find all values of \(m\) such that \( y = x^{m} \) is a solution of \( 2x^2y'' + 3xy' - 15y = 0 \).
\(x^2x^{m-2}=x^m\) and \(x\,x^{m-1}=x^m\). That is the whole reason \(x^m\) is the right guess for this equation (it is a Cauchy–Euler equation). So the left side is \(x^{m}\big[2m(m-1) + 3m - 15\big]\).
STEP 4 — Set the bracket to zero
\(2m^2 - 2m + 3m - 15 = 2m^2 + m - 15 = 0\), which factors as \((2m-5)(m+3)=0\).
STEP 5 — Solve and check
\(\boxed{m = \tfrac52 \text{ or } m = -3}\). Check: \(2(\tfrac{25}{4}) + \tfrac52 - 15 = 12.5+2.5-15 = 0\) ✓ and \(2(9) - 3 - 15 = 0\) ✓.
Note
\(x^{m}\) is only a solution for \(x>0\) when \(m=\tfrac52\), and for \(x\neq0\) when \(m=-3\). §1.2 problem 2 builds the general solution \(y=c_1x^{5/2}+c_2x^{-3}\) from these two roots.
6
\( y = \sin(x) \) is an explicit solution of the first-order equation \( \dfrac{dy}{dx} = \sqrt{1-y^2} \). Find the interval \(I\) of definition. (Hint: \(I\) is not \((-\infty,\infty)\).)
Solution
STEP 1 — Compute the left side
\(\dfrac{dy}{dx} = \cos x\).
STEP 2 — Compute the right side
\(\sqrt{1-\sin^2 x} = \sqrt{\cos^2 x} = |\cos x|\) — the square root symbol means the non-negative root, always.
STEP 3 — Set them equal
The DE holds exactly where \(\cos x = |\cos x|\), i.e. where \(\cos x \ge 0\).
STEP 4 — Solve the inequality
\(\cos x \ge 0\) on \(\big[-\tfrac{\pi}{2} + 2k\pi,\; \tfrac{\pi}{2} + 2k\pi\big]\). Take the interval around the origin: \(\big[-\tfrac{\pi}{2},\tfrac{\pi}{2}\big]\).
STEP 5 — Trim the endpoints
At \(x=\pm\tfrac{\pi}{2}\) the right side is \(0\) and so is \(\cos x\), so the equation technically still holds — but the standard interval of definition is taken open so the solution is interior to where the DE is well behaved:
On \((\tfrac{\pi}{2}, \tfrac{3\pi}{2})\), \(\cos x\) is negative while \(\sqrt{1-y^2}\) is still \(\ge 0\) — so \(\sin x\) satisfies \(y' = -\sqrt{1-y^2}\) there, a different equation. A formula being defined everywhere does not make it a solution everywhere.
⌕Source & references
Prof. Williams — Ch. 1.1 Definitions and Terminology, Write-Up, and Differential Equations with Boundary-Value Problems — Lecture Notes, §1.1.
Zill, Differential Equations with Boundary-Value Problems — §1.1, the course text.
A DE hands you a family of solutions. Initial conditions pick one out of the family — when they can. This section is about when that works, and the theorem that says so.
①What an IVP is
An initial value problem is a DE together with conditions specified at a single point. For an \(n\)th-order equation you get \(n\) of them — values for \(y, y', \dots, y^{(n-1)}\), all at the same \(x_0\).
Initial conditions let us single out one solution from a family. It is possible for more than one solution in a family to share the same initial conditions, which raises the two questions the section exists to answer:
Are there solutions? — the question of existence.
Is there more than one? — the question of uniqueness.
②The Existence & Uniqueness Theorem
As stated in the course
Let \(R\) be a rectangular region in the \(xy\)-plane defined by \(x\in[a,b]\), \(y\in[c,d]\) that contains the point \((x_0,y_0)\) in its interior. If \(f(x,y)\) and \(\dfrac{\partial f}{\partial y}\) are continuous on \(R\), then there exists an interval \(I_0=(x_0-h,\;x_0+h)\), \(h>0\), contained in \([a,b]\), and a unique function \(y(x)\) defined on \(I_0\) that solves the IVP \(\dfrac{dy}{dx}=f(x,y)\) with \(y(x_0)=y_0\).
How to actually use it
Three mechanical steps, every time. (1) Solve for \(y'\) so the equation reads \(y'=f(x,y)\) — the theorem is about that form and no other. (2) Compute \(\partial f/\partial y\). (3) Ask whether both are continuous on some rectangle with the given point inside it, not merely on its edge.
One-way theorem
Failing the hypotheses does not prove there is no unique solution — it only means the theorem is silent. The correct wording on an exam is "the theorem does not guarantee," never "there is no unique solution."
⌖ Worked example 1 — interval of existence
\(y=\dfrac{1}{x^2+c}\) solves \(y' + 2xy^2 = 0\). Find the solution with \(y(-2)=\tfrac12\) and its interval of existence.
The formula fails when \(x^2-2=0\), i.e. \(x=\pm\sqrt2\).
STEP 4 — Pick the branch containing \(x_0\)
The initial point is \(x=-2\), which lies to the left of \(-\sqrt2\approx-1.414\). The interval of existence is \(\big(-\infty,\,-\sqrt2\,\big)\) — not all three branches, and not the whole line.
⌖ Worked example 2 — conditions from a tangent line
Find \(y=f(x)\) with \(y''=12x-2\) at each point, given that \(y=-x+5\) is tangent to the graph at \(x=1\).
STEP 1 — Decode "tangent at \(x=1\)"
Tangency gives you two conditions, not one: the slope matches, \(y'(1)=-1\), and the point is shared, \(y(1)=-1+5=4\).
\(4 = 2-1-5+c_2 \Rightarrow c_2=8\), so \(y = 2x^3-x^2-5x+8\).
⌖ Worked example 3 — a region of uniqueness
Determine a region of the \(xy\)-plane on which \(y'-y=x\) has a unique solution through \((x_0,y_0)\).
STEP 1 — Solve for \(y'\)
\(y' = y + x\), so \(f(x,y)=x+y\).
STEP 2 — Continuity of \(f\)
A polynomial in \(x\) and \(y\) — continuous on all of \(\mathbb{R}^2\).
STEP 3 — Continuity of \(\partial f/\partial y\)
\(f_y = 1\), the constant function — also continuous everywhere.
STEP 4 — Conclude
Both hypotheses hold on every rectangle, so there is a unique solution through any point of the plane.
✎§1.2 Classwork — all five, worked
1
Find \(c_1\) and \(c_2\) so that \( y = c_1e^{-3x} + c_2e^{2x} \) solves \( y'' + y' - 6y = 0 \) with \( y(0)=1,\; y'(0)=-1 \).
Solution
STEP 1 — Confirm it really is a solution
Substituting \(y=e^{rx}\) gives \(r^2+r-6=(r+3)(r-2)=0\), roots \(r=-3,2\) — exactly the two exponents given. (Contrast §1.1 problem 3, where the signs were swapped and the answer was "no.")
STEP 2 — First condition
\(y(0)=c_1e^{0}+c_2e^{0}=c_1+c_2=1\). — call this (A).
STEP 3 — Differentiate, then second condition
\(y' = -3c_1e^{-3x} + 2c_2e^{2x}\), so \(y'(0) = -3c_1 + 2c_2 = -1\). — call this (B).
STEP 4 — Substitute (A) into (B)
From (A), \(c_1 = 1-c_2\). Then \(-3(1-c_2)+2c_2 = -1 \Rightarrow -3+3c_2+2c_2=-1 \Rightarrow 5c_2 = 2\).
STEP 5 — Back-substitute
\(c_2 = \tfrac25\), so \(c_1 = 1-\tfrac25 = \tfrac35\).
Find \(m\) so that \( y=x^{m} \) solves \( 2x^2y'' + 3xy' - 15y = 0 \) with \( y(1)=2,\; y'(1)=1 \).
Solution
STEP 1 — Get the exponents (§1.1 #5)
Substituting \(y=x^m\) gives \(x^m\big[2m(m-1)+3m-15\big]=x^m(2m^2+m-15)=0\), so \((2m-5)(m+3)=0\) and \(m=\tfrac52\) or \(m=-3\).
STEP 2 — Two roots, two conditions
A single \(x^m\) can't satisfy two independent conditions. Because the equation is linear and homogeneous, superposition applies: build \(y = c_1x^{5/2} + c_2x^{-3}\).
\(y'(1) = \tfrac52\cdot\tfrac{14}{11} - 3\cdot\tfrac{8}{11} = \tfrac{35}{11}-\tfrac{24}{11}=1\) ✓. Both \(x^{5/2}\) and \(x^{-3}\) require \(x>0\), and the initial point is \(x=1\), so the interval of existence is \((0,\infty)\).
3
Does the existence-and-uniqueness theorem guarantee that \( 2xy' = \sqrt{xy^2-9} \) has a unique solution at the point \((1,3)\)?
Solution
STEP 1 — Put it in \(y'=f(x,y)\) form
\(f(x,y) = \dfrac{\sqrt{xy^2-9}}{2x}\). The theorem says nothing about the undivided form, so this step is mandatory.
STEP 2 — Evaluate the radicand at the point
At \((1,3)\): \(xy^2-9 = (1)(9)-9 = 0\). The point sits exactly on the boundary of the region where \(f\) is defined.
STEP 3 — Is \(f\) continuous on a rectangle around \((1,3)\)?
No. Any rectangle containing \((1,3)\) in its interior also contains points with \(y\) slightly less than 3, where \(xy^2-9<0\) and \(f\) is not even real-valued. The first hypothesis already fails.
STEP 4 — Check \(\partial f/\partial y\) too
\(\dfrac{\partial f}{\partial y} = \dfrac{1}{2x}\cdot\dfrac{2xy}{2\sqrt{xy^2-9}} = \dfrac{y}{2\sqrt{xy^2-9}}\). At \((1,3)\) the denominator is \(0\), so it blows up — the second hypothesis fails as well.
STEP 5 — State the conclusion carefully
The theorem does not guarantee a unique solution at \((1,3)\). Both hypotheses fail there. Note what this does not say: the theorem is a sufficient condition, so its failure leaves the question open rather than answering it "no."
Where it would work
At a point with \(xy^2>9\) and \(x\neq0\) — say \((1,4)\) — both \(f\) and \(f_y\) are continuous on a small rectangle, and the theorem does apply.
4
Find \(y=f(x)\) whose graph has slope \(5e^{3x}-4x\) at each point \((x,y)\) and \(y\)-intercept \((0,3)\).
Solution
STEP 1 — Translate the words into an IVP
"Slope at each point" is \(y'\); "\(y\)-intercept \((0,3)\)" is \(y(0)=3\). So: \(\dfrac{dy}{dx} = 5e^{3x}-4x,\; y(0)=3\).
STEP 2 — Notice it is trivially separable
The right side has no \(y\) in it, so you can integrate straight across — no method needed beyond antidifferentiation.
STEP 3 — Integrate
\(y = \displaystyle\int\!\big(5e^{3x}-4x\big)dx = \tfrac53 e^{3x} - 2x^2 + C\). The \(\tfrac13\) comes from the chain rule on \(e^{3x}\) — the most common slip here.
STEP 4 — Apply the initial condition
\(y(0) = \tfrac53 e^{0} - 0 + C = \tfrac53 + C = 3 \Rightarrow C = \tfrac43\).
STEP 5 — Answer
\(\boxed{\;y = \tfrac53 e^{3x} - 2x^2 + \tfrac43\;}\), valid on \((-\infty,\infty)\) — nothing was divided by, so there is no singular point.
5
Two functions \(f\) and \(g\) are graphed through the point \((0,\tfrac12)\): \(g\) is increasing there, \(f\) is decreasing. Which is a plausible solution of the IVP \( y' = x-2y,\; y(0)=\tfrac12 \)?
Solution
STEP 1 — You do not need to solve anything
The DE evaluates the slope directly at any point. Use it as an oracle at the one point both curves share.
\(-1 < 0\), so the solution curve must be decreasing as it passes through \((0,\tfrac12)\) — and with slope exactly \(-1\), i.e. at \(45^\circ\) downward.
STEP 4 — Eliminate
\(g\) rises through the point, so \(g\) is impossible. \(f\) falls through it.
STEP 5 — Answer
\(f\) is the plausible solution curve.
Worth carrying forward
This "evaluate the DE at the given point and read the sign" move is the entire idea of §2.1 — a direction field is just this done at every point of a grid at once.
⌕Source & references
Prof. Williams — Lecture Notes, §1.2 Initial-Value Problems (all three worked examples above are from the notes).
Zill — §1.2, including the statement of the existence & uniqueness theorem quoted on the classwork sheet.
Lesson L2 of the 3-week track covers the same theorem with a different set of examples.
Math 275 · §1.3 Prof. Williams
DEs as Mathematical Models
Ten standard models you are expected to be able to write down cold, and the habit of building one from scratch: name the variables, choose the level of resolution, state the assumptions, then translate each assumption into a term.
①How a model gets built
Construction starts with figuring out which variables are responsible for the change you observe. You deliberately do not include all of them at first — you keep a simple general model that can be refined by adding the others later. That choice is called specifying the level of resolution of the model.
Next you commit to a set of reasonable assumptions, including any formulas applicable to the system. This course revisits the same application problems repeatedly at increasing resolution — the tank you drain in §1.3 comes back in §3.2 with more physics in it.
The one sentence that does most of the work
"The rate at which \(X\) changes is proportional to \(Y\)" becomes \(\dfrac{dX}{dt}=kY\). Nearly every model below is that sentence with different nouns — and the sign of \(k\) is where the physics lives.
②The ten models
Population growth
\[ \frac{dP}{dt}=kP \]\(P(t)\) is population at time \(t\); \(k>0\) is the constant of proportionality.
Radioactive decay
\[ \frac{dA}{dt}=kA \]\(A(t)\) is the amount of substance at time \(t\); \(k<0\). Same equation as growth — only the sign of \(k\) differs.
Newton's law of cooling
\[ \frac{dT}{dt}=k\,(T-T_m) \]\(T(t)\) is the temperature of the object, \(T_m\) the temperature of the surrounding medium. If \(T_m\) is constant then \(k<0\).
Spread of disease
\[ \frac{dx}{dt}=kxy \]\(x(t)\) is the number of diseased people, \(y(t)\) the number not diseased; the product \(xy\) counts the interactions between the two groups.
Chemical reactions
1st order (decomposition): \[ \frac{dX}{dt}=kX,\qquad k<0 \]
2nd order (\(A\) and \(B\) combine to form \(C\)): \[ \frac{dX}{dt}=k(\alpha-X)(\beta-X) \]
\(X(t)\) is the amount of \(C\) created; \(\alpha-X\) is the amount of \(A\) still unconverted, \(\beta-X\) the amount of \(B\).
Mixtures
\[ \frac{dA}{dt}=R_{\text{in}}-R_{\text{out}} \]
\(A(t)\) is the amount of substance in the mixture. \(R_{\text{in}} = (\text{concentration in inflow})\times(\text{input rate of mixture})\), and \(R_{\text{out}} = (\text{concentration in outflow})\times(\text{output rate of mixture})\).
Draining a tank (Torricelli)
\[ \frac{dV}{dt}=-A_h\sqrt{2gh}, \qquad \frac{dh}{dt}=-\frac{A_h}{A_w}\sqrt{2gh} \]
\(v=\sqrt{2gh}\) is the speed of water leaving, \(A_h\) the area of the hole, \(A_w\) the top surface area of the water, \(h(t)\) the height.
Series circuits
\[ i=\frac{dq}{dt},\qquad E(t)=L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac1C q \]
Voltage across the inductor \(=L\frac{di}{dt}\), across the resistor \(=iR\), across the capacitor \(=\frac1Cq\).
Falling bodies
\[ F=ma=m\frac{d^2s}{dt^2}=-mg \;\Longrightarrow\; \frac{d^2s}{dt^2}=-g \]
\(s(t)\) position, \(v=\frac{ds}{dt}\), \(a=\frac{dv}{dt}=\frac{d^2s}{dt^2}\); \(W=mg\) is the weight.
Air resistance
\[ m\frac{dv}{dt}=mg-kv \]\(-kv\) is the viscous damping; \(mg\) is the weight.
Suspended cables
\[ \tan\theta=\frac{|T_2|\sin\theta}{|T_2|\cos\theta}=\frac{|W|}{|T_1|}=\frac{dy}{dx} \]
\(W\) is the total vertical load between \(P_1\) (the lowest point of the cable) and \(P_2\); \(T_1,T_2\) are tangent vectors with \(|T_1|=|T_2|\cos\theta\) and \(|W|=|T_2|\sin\theta\).
✎§1.3 Classwork — all three, worked
1
A person \(P\) starts at the origin and moves along the positive \(x\)-axis, pulling a weight along a curve \(C\) (a tractrix). The weight starts on the \(y\)-axis at \((0,s)\) and is pulled by a taut rope of constant length \(s\). Determine a DE for the path \(C\), assuming the rope is always tangent to \(C\).
Solution
STEP 1 — Name what the rope is
The rope runs from the weight at \((x,y)\) on the curve to the person \(P\) on the \(x\)-axis. "Always tangent to \(C\)" means the rope's slope is \(dy/dx\).
STEP 2 — Note the direction
\(y\) decreases as \(P\) moves in the \(+x\) direction, so the slope is negative: \(y' = -\tan\theta\) for the acute angle \(\theta\) the rope makes with the vertical.
STEP 3 — Get the horizontal leg from Pythagoras
The rope is the hypotenuse, length \(s\); the vertical leg from the weight down to the \(x\)-axis has length \(y\). So the horizontal distance from the weight to \(P\) is \(d(x,P)=\sqrt{s^2-y^2}\).
STEP 4 — Form the slope
Rise over run, with the sign from Step 2: \(\dfrac{dy}{dx} = -\dfrac{y}{\sqrt{s^2-y^2}}\).
STEP 5 — Why write it this way
The angle \(\theta\) changes as \(P\) moves, so an answer in terms of \(\theta\) is useless. Trading it for \(y\) and the given constant \(s\) is the whole point of Step 3.
\(\boxed{\;\dfrac{dy}{dx} = -\dfrac{y}{\sqrt{s^2-y^2}}\;}\) on \(0<y<s\).
2
Virga — raindrops that evaporate before reaching the ground. A spherical drop of radius \(r_0\) falls from rest at time \(t_0\) and evaporates. (a) Assuming the drop stays spherical and loses mass at a rate proportional to its surface area, show that \(r\) decreases at a constant rate, and find \(r(t)\). (b) With down as positive and air resistance ignored, model the velocity \(v\). (Hint: with \(m\) changing, Newton's second law is \(F=\frac{d}{dt}(mv)\).)
Solution
STEP 1 — (a) Write mass in terms of radius
Volume of a sphere \(V=\tfrac43\pi r^3\), and mass is some constant multiple of volume: \(m = kV = k\tfrac43\pi r^3\).
STEP 2 — Differentiate with the chain rule
\(m' = k\,4\pi r^2\,r'\).
STEP 3 — Impose the assumption
Surface area is \(S=4\pi r^2\), and the assumption says \(m' = \lambda\,S\) for some constant \(\lambda\). Comparing with Step 2, \(\lambda = k r'\) — a constant. Since \(k\) is constant, \(r'\) is constant. That is what was to be shown.
STEP 4 — Integrate for \(r(t)\)
\(r' = k \Rightarrow r = \int k\,dt = kt + c\). At \(t=0\), \(r=r_0\), so \(c=r_0\):
\(\boxed{\;r(t) = kt + r_0\;}\) with \(k<0\) (the drop shrinks).
STEP 5 — (b) Apply the momentum form of Newton's law
\(\dfrac{d}{dt}(mv) = mg\). Expand with the product rule: \(mv' + vm' = mg\).
STEP 6 — Substitute \(m\) and \(m'\)
\(k\tfrac43\pi r^3\,v' + v\,k\,4\pi r^2 = k\tfrac43\pi r^3 g\). Divide through by \(k\tfrac43\pi r^3\):
A first-order linear equation — you will solve it by integrating factor in §2.3.
3
A plane curve \(C\) is revolved about the \(x\)-axis, generating a surface with the property that every light ray \(L\) parallel to the \(x\)-axis is reflected to the origin \(O\). Using "angle of incidence = angle of reflection," find a DE for \(C\). (Hint: the figure gives \(\phi = 2\theta\). Why?)
Solution
STEP 1 — Answer the hint's "why"
\(\theta\) is the angle the tangent line makes with the horizontal; the incoming ray is horizontal, so the angle between the ray and the tangent is also \(\theta\), and the reflection law doubles it. The reflected ray therefore makes angle \(\phi = 2\theta\) with the horizontal — the exterior-angle argument in the figure.
STEP 2 — Name both tangents
The reflected ray runs from the point \((x,y)\) to the origin, so \(\tan\phi = \dfrac{y}{x}\). The tangent line to \(C\) has \(\tan\theta = \dfrac{dy}{dx}\). Write \(m = dy/dx\) for short.
STEP 3 — Use the double-angle identity
\(\tan\phi = \tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}\), so \(\dfrac{y}{x} = \dfrac{2m}{1-m^2}\).
STEP 4 — Clear denominators into a quadratic in \(m\)
This is a homogeneous equation (solvable by \(y=vx\) in §2.5); its solutions are the parabolas \(y^2 = 4a(x+a)\) with focus at the origin. That is why the shape shows up in telescopes, satellite dishes, headlights, and solar collectors — exactly the applications the problem lists.
⌕Source & references
Prof. Williams — Ch. 1.3 Differential Equations as Mathematical Models, Write-Up (the ten models and problems 1–2 are worked there); Lecture Notes §1.3.
Zill — §1.3 and its exercise set.
These models return in Chapter 3 (Linear and Non-Linear Models) at higher resolution — the tank, the circuit, and the falling body all come back.
Math 275 · §2.1 Prof. Williams
Solution Curves Without a Solution
Nothing in this section gets solved. Both tools — direction fields and phase lines — read an equation's behaviour straight off its geometry: where the tangent marks point, and which constant solutions pull neighbours in or push them away.
①Direction fields — for any \(y'=f(x,y)\)
There will be times when no explicit solution is available. In those cases graphs let us analyse the system anyway. A direction field is built by plugging points \((x,y)\) into the DE; the result is the slope of the solution curve through that point. Draw a short segment with that slope at each grid point and you have a "flow map."
⌖ From the lecture notes
Consider \(y' = x(2-y)\) and sample a few points:
(0, 1)
\(y' = 0\cdot(2-1)=0\) — draw a short horizontal mark. Because the rate of change is 0 there, \((0,1)\) is an equilibrium point (also called a critical point or stability point).
(1, 1)
\(y' = 1\cdot(2-1)=1\) — a mark at \(45^\circ\).
(1, 2)
\(y' = 1\cdot(2-2)=0\) — horizontal again.
Reading the result
A direction field behaves like the current of a river: pick a starting point and follow the marks to trace the solution through it. Here every point with \(y=2\) has slope 0, so the constant function \(y=2\) is itself a solution — an equilibrium solution.
Two structural facts worth stating on any sketch question:
An isocline is the curve \(f(x,y)=c\), where every mark has the same slope \(c\). Setting \(c=0\) gives the nullcline, where solutions level off.
Solution curves never cross — where two met, the field would have to specify two slopes at one point.
②Phase lines — only for autonomous \(y'=f(y)\)
An autonomous DE is one in which the independent variable does not appear. When \(x\) is absent from the right side, the slope depends on height alone, so the entire picture collapses onto a single vertical \(y\)-axis. In an autonomous DE an equilibrium point is an equilibrium solution.
The four-step routine
1. Find the critical points. Solve \(f(y)=0\). Each root \(c\) gives the constant solution \(y\equiv c\) — a horizontal line in the \(xy\)-plane. 2. Sign-chart \(f(y)\) between them. One test value per interval. \(f>0\) → increasing there; \(f<0\) → decreasing. 3. Draw the arrows. Up where \(f>0\), down where \(f<0\). Solutions cannot cross the critical lines, so each is trapped in its strip and is monotone. 4. Classify by what both neighbours do. In from both sides → attractor (stable). Out on both sides → repeller (unstable). In on one side, out on the other → semi-stable.
Fast second opinion
The derivative test: \(f'(c)<0\) → attractor, \(f'(c)>0\) → repeller, \(f'(c)=0\) → look harder. Semi-stable only happens at a repeated root of \(f\), so distinct roots rule it out immediately.
Example from the notes: \(\dfrac{dy}{dx} = -3y+y^2\) has critical points \(y=0\) and \(y=3\); \(y=3\) is unstable and \(y=0\) is stable.
✎§2.1 Classwork — all five, worked
1
On the printed direction field, draw the solution curve satisfying \( y(-2) = -4 \).
Solution
STEP 1 — Describe the field before drawing
Every mark points up and to the right, so the slope is positive everywhere and every solution is increasing — your curve can never turn back down. On the left half (\(x<0\)) with large \(|y|\) the marks are near-vertical; on the right half (\(x>0\)) with large \(|y|\) they flatten toward 0. Along both axes the marks sit at \(45^\circ\).
STEP 2 — Start at the point
Mark \((-2,-4)\) in the third quadrant. The marks there lean up-right, a bit steeper than \(45^\circ\). Set the pencil down and follow them.
STEP 3 — Trace to the right
The curve climbs, crosses the \(x\)-axis near \(x\approx1.8\), then bends over as it enters the region of flat marks, settling toward \(y\approx4.5\) at the right edge. Concave down on this stretch.
STEP 4 — Trace to the left
Moving left, the marks stand up fast. The curve drops steeply and runs off the bottom of the window by about \(x\approx-3.6\) — nearly vertical, not a gentle tail.
STEP 5 — The answer as a description
A single increasing curve: steeply rising out of the bottom-left, through \((-2,-4)\), crossing the \(x\)-axis around \(x\approx1.8\), then flattening into a near-horizontal shelf near \(y\approx4\)–\(5\) as it goes right.
Why the shape
The printed field is \(dy/dx = e^{-0.01xy^2}\) (Zill Fig. 2.1.13). You aren't given it and don't need it — but the exponent is positive when \(x<0\) and negative when \(x>0\), which is exactly the vertical-then-flat behaviour.
Easy to miss
Graders check two things: the curve must be tangent to the marks it passes through, not merely near them, and it must be monotone increasing across the whole window. A sketch that peaks and comes back down contradicts a field with no horizontal marks in it.
2
For \( \dfrac{dy}{dx} = y(x-2)^2 - 5 \), describe the slopes of the lineal elements along the lines \(x=3\), \(x=2\), and \(x=0\).
Solution
STEP 1 — What "along the line \(x=k\)" means
Substitute that number for \(x\) and read what is left as a function of \(y\) alone. The factor \((x-2)^2\) becomes a constant, so each vertical line gives a linear slope rule in \(y\).
STEP 2 — \(x=2\)
\(\dfrac{dy}{dx} = y(0)^2-5 = -5\). The \(y\) drops out entirely: every element along \(x=2\) has slope exactly \(-5\), no matter how high or low you look — a column of identical, steeply falling parallel marks. This line is the isocline for \(c=-5\).
STEP 3 — \(x=3\)
\(\dfrac{dy}{dx} = y(1)^2 - 5 = y-5\). Slopes climb one-for-one with \(y\): \(-5\) at \(y=0\), horizontal at \(y=5\), positive above, negative below — a fan rotating steadily counter-clockwise as you move up the line.
STEP 4 — \(x=0\)
\(\dfrac{dy}{dx} = y(-2)^2 - 5 = 4y-5\). Same shape as \(x=3\) but four times as sensitive: horizontal at \(y=\tfrac54\), and by \(y=5\) the slope is already \(15\) — essentially vertical.
STEP 5 — Tie it together
The general isocline for slope \(c\) is \(y=\dfrac{c+5}{(x-2)^2}\). Two consequences worth stating: the nullcline (\(c=0\)) is \(y=\dfrac{5}{(x-2)^2}\), and the entire \(x\)-axis has slope \(-5\), since \(y=0\) kills the first term for every \(x\).
Easy to miss
The distances from 2 are squared: \((3-2)^2=1\) but \((0-2)^2=4\). Writing "slope \(=y-5\)" for \(x=0\) by forgetting to square the \(-2\) is the standard slip.
3
For \( \dfrac{dy}{dx} = y-y^2 \) with \(y(0)=y_0\), sketch a typical solution when (a) \(y_0>1\), (b) \(y_0\in(0,1)\), (c) \(y_0<0\).
Solution
STEP 1 — Critical points
\(y-y^2 = y(1-y) = 0 \Rightarrow y=0,\; y=1\). These are the two constant solutions, and they slice the plane into exactly the three strips the problem's three cases correspond to.
STEP 2 — Sign chart
\(y>1\): test \(y=2 \to 2-4=-2\), decreasing. \(0<y<1\): test \(y=\tfrac12 \to \tfrac12-\tfrac14=+\tfrac14\), increasing. \(y<0\): test \(y=-1 \to -1-1=-2\), decreasing.
So \(y=1\) is an attractor and \(y=0\) is a repeller.
STEP 3 — Concavity
Differentiate the equation itself rather than any solution formula: \(y'' = (1-2y)\,y'\). The only place concavity can flip is \(y=\tfrac12\) — which matters for case (b) and nowhere else.
STEP 4 — (a) \(y_0>1\)
Decreasing and concave up, falling toward the line \(y=1\) and approaching it asymptotically as \(x\to\infty\). Backward in \(x\) it climbs and blows up to \(+\infty\) at a finite \(x\). It never touches \(y=1\).
STEP 5 — (b) \(y_0\in(0,1)\)
The classic logistic S-curve: increasing, trapped between the two horizontal lines, rising from \(y\to0\) as \(x\to-\infty\) to \(y\to1\) as \(x\to+\infty\), with an inflection where it crosses \(y=\tfrac12\) — concave up below it, concave down above.
STEP 6 — (c) \(y_0<0\)
Decreasing and concave down, pushed away from \(y=0\). Backward in \(x\) it rises toward \(y=0\) from below; forward it accelerates to \(-\infty\) at a finite \(x\) — a vertical asymptote, not a slow drift.
Confirmation
The closed form is \(y = \dfrac{y_0e^{x}}{1-y_0+y_0e^{x}}\), whose denominator vanishes at \(x=\ln\!\big(1-\tfrac{1}{y_0}\big)\) — positive when \(y_0<0\), negative when \(y_0>1\). That is exactly where each blow-up sits.
Easy to miss
Curves in the three strips are separate solutions, not one continuous curve. Uniqueness forbids crossing \(y=0\) and \(y=1\), so a sketch that starts below 0 and rises through both lines is wrong no matter how good it looks.
4
Classify the critical points of \( \dfrac{dy}{dx} = y-y^2+2 \) as stable (attractor), unstable (repeller), or semi-stable.
Solution
STEP 1 — Factor, minus sign first
\(y - y^2 + 2 = -(y^2-y-2) = -(y-2)(y+1)\). Critical points \(y=-1\) and \(y=2\). Pulling the minus out before factoring is where sign errors get prevented.
STEP 2 — Rule out semi-stable immediately
Two distinct roots, so neither can be semi-stable — that verdict requires a repeated factor.
STEP 3 — Sign chart
\(y>2\): test \(y=3 \to -(1)(4)=-4\), decreasing. \(-1<y<2\): test \(y=0 \to -(-2)(1)=+2\), increasing. \(y<-1\): test \(y=-2 \to -(-4)(-1)=-4\), decreasing.
STEP 4 — Read the arrows
Below \(y=-1\) solutions fall away; just above it they rise away. Both neighbours flee → repeller. Below \(y=2\) solutions rise toward it; above it they fall toward it. Both neighbours approach → attractor.
STEP 5 — Answer & check
\(y=-1\) unstable (repeller); \(y=2\) asymptotically stable (attractor); neither is semi-stable. Derivative test: \(f'(y)=1-2y\), so \(f'(-1)=3>0\) (repeller) and \(f'(2)=-3<0\) (attractor). ✓
Easy to miss
Because the \(y^2\) term is negative the parabola opens downward, so the larger root is the attractor. With a positive leading coefficient it is the other way round. Factoring before sign-charting is what keeps this straight.
5
Sketch the one-dimensional phase portrait for \( y' = y^2 - by \), where \(b\in\mathbb{R}^-\).
Solution
STEP 1 — Factor
\(y^2 - by = y(y-b) = 0 \Rightarrow y=0,\; y=b\).
STEP 2 — Order them on the line
This is the step the condition \(b\in\mathbb{R}^-\) exists for. Since \(b\) is strictly negative, \(b\) sits below \(0\) on the phase line — and the two points stay distinct, so no semi-stable case arises.
STEP 3 — Sign chart
\(y>0\): \(y\) is \(+\), \((y-b)\) is \(+\) → product \(+\), increasing. \(b<y<0\): \(y\) is \(-\), \((y-b)\) is \(+\) → \(-\), decreasing. \(y<b\): both \(-\) → \(+\), increasing.
STEP 4 — Classify
Below \(b\) the arrows point up toward it; between \(b\) and \(0\) they point down toward it — \(y=b\) is an attractor. Just below 0 the arrows point down, away; above 0 they point up, away — \(y=0\) is a repeller.
STEP 5 — The portrait
A vertical line with two marked points, \(b\) below \(0\): ↑ below \(y=b\) · \(y=b\) attractor · ↓ between · \(y=0\) repeller · ↑ above \(y=0\). Equivalently \(f'(y)=2y-b\), so \(f'(b)=b<0\) (attractor) and \(f'(0)=-b>0\) (repeller) — both signs coming straight from \(b<0\).
Easy to miss
Don't assume \(b\) is above 0 out of habit. If \(b\) were positive the order flips and so do the labels: 0 becomes the attractor and \(b\) the repeller. At \(b=0\) the roots merge into a single semi-stable point — which is precisely why the problem excludes it.
✔Answers at a glance
1. Increasing curve through \((-2,-4)\): near-vertical rise on the left, crosses \(y=0\) near \(x\approx1.8\), flattens near \(y\approx4\)–\(5\) on the right.
Say why, not just what. "\(y=2\) is an attractor" earns less than "\(f>0\) below it and \(f<0\) above it, so both neighbours approach." The sign chart is the argument. Factor before you sign-chart — problems 4 and 5 both hinge on it. Respect the barriers. Constant solutions cannot be crossed; every sketch in problem 3 stays inside one strip. Distinguish "approaches" from "reaches." Attractors are approached asymptotically and never touched; blow-ups happen at finite \(x\).
⌕Source & references
Prof. Williams — Lecture Notes §2.1 Solution Curves Without a Solution (direction fields, autonomous DEs, the three phase-portrait figures).
Zill — §2.1, including Fig. 2.1.13 used in problem 1.
Lesson L2 of the 3-week track covers slope fields and equilibria from a different angle.
Math 275 · §2.2 Prof. Williams
Separable Variables
The first section where you actually solve something. If the right side factors into a piece that depends only on \(x\) times a piece that depends only on \(y\), you can push each variable to its own side of the equation and integrate.
①The method
If the equation has the form \(\dfrac{dy}{dx} = \dfrac{g(x)}{f(y)}\), separate it:
\[ f(y)\,dy = g(x)\,dx \;\Longrightarrow\; \int h(y)\,dy = \int g(x)\,dx \;\Longrightarrow\; H(y) = G(x) + c \]
where \(h(y)=\dfrac{1}{f(y)}\) as needed. Then solve for \(y(x)\) if you can.
Explicit vs. implicit
If you can isolate \(y\), the result is an explicit solution. If you cannot — and often you cannot — the relation \(H(y)=G(x)+c\) is a perfectly valid implicit solution. Do not mangle algebra trying to force an explicit form that does not exist.
One constant, not two
Integrating both sides technically gives \(c_1\) and \(c_2\); they immediately merge into a single \(c = c_2-c_1\). Write one constant, on the side that keeps the algebra cleanest.
Losing a solution
Dividing by \(f(y)\) throws away every root of \(f(y)=0\). Those constant functions are usually genuine singular solutions not contained in the general family — check them separately and mention them.
⌖ Worked example 1 — implicit answer (from the notes)
Solve \( u' = \dfrac{1+t^4}{ut^2 + u^4t^2} \).
STEP 1 — Factor the denominator
\(ut^2+u^4t^2 = t^2(u+u^4)\), so \(\dfrac{du}{dt} = \dfrac{1+t^4}{t^2(u+u^4)}\) — now visibly separable.
STEP 2 — Separate
\((u+u^4)\,du = \dfrac{1+t^4}{t^2}\,dt = \big(t^{-2}+t^2\big)dt\). Splitting the fraction first is what makes the right integral trivial.
\(\dfrac{1}{y-3} = x - c\), so \(y - 3 = \dfrac{1}{x-c}\) and
\(\boxed{\;y = 3 + \dfrac{1}{x-c}\;}\)
STEP 5 — The lost solution
\(y \equiv 3\) makes \((y-3)^2=0\) and \(dy=0\), so it satisfies the equation — but no finite \(c\) produces it. It is a singular solution and belongs in the answer.
Check
\(y' = -\dfrac{1}{(x-c)^2}\) and \((y-3)^2 = \dfrac{1}{(x-c)^2}\); they sum to 0 ✓.
\(\int y^{-3}dy = -\dfrac{1}{2y^2}\) and \(\int -3x\,dx = -\dfrac{3x^2}{2}\), so \(-\dfrac{1}{2y^2} = -\dfrac{3x^2}{2} + c\).
STEP 3 — Tidy the signs
Multiply by \(-2\): \(\dfrac{1}{y^2} = 3x^2 + C\) with \(C=-2c\). Doing this before applying the initial condition avoids a fraction fight.
STEP 4 — Apply \(y(1)=1\)
\(\dfrac{1}{1} = 3(1)^2 + C \Rightarrow C = -2\), so \(\dfrac{1}{y^2} = 3x^2-2\).
STEP 5 — Solve explicitly and pick the branch
\(y^2 = \dfrac{1}{3x^2-2}\). Since \(y(1)=1>0\), take the positive root:
\(\boxed{\;y = \dfrac{1}{\sqrt{3x^2-2}}\;}\)
STEP 6 — Interval of existence
Need \(3x^2-2>0\), i.e. \(|x|>\sqrt{2/3}\approx0.816\). The initial point \(x=1\) sits in the right branch, so \(I = \big(\sqrt{2/3},\,\infty\big)\).
Check
\(y=(3x^2-2)^{-1/2} \Rightarrow y' = -3x(3x^2-2)^{-3/2}\), and \(3xy^3 = 3x(3x^2-2)^{-3/2}\). Sum \(=0\) ✓. (Also \(y\equiv0\) is a singular solution of the DE, excluded here by the initial condition.)
The \(dx\) term carries only \(y\)'s and the \(dy\) term only \(x\)'s. Divide by the product \((e^{y}+1)^2e^{-y}\,(e^{x}+1)^3e^{-x}\) to un-cross them.
STEP 2 — Separate
\(\dfrac{dx}{(e^{x}+1)^3e^{-x}} = -\dfrac{dy}{(e^{y}+1)^2e^{-y}}\), i.e. \(\dfrac{e^{x}}{(e^{x}+1)^3}\,dx = -\dfrac{e^{y}}{(e^{y}+1)^2}\,dy\), using \(1/e^{-x}=e^{x}\).
\(d\!\left[\tfrac12(e^{x}+1)^{-2}\right] = -e^{x}(e^{x}+1)^{-3}dx\) and \(d\!\left[(e^{y}+1)^{-1}\right] = -e^{y}(e^{y}+1)^{-2}dy\). Setting the sum to zero and multiplying through by \(-(e^{x}+1)^3(e^{y}+1)^2 e^{-x-y}\) returns the original equation ✓.
5
Find an explicit solution to \( (1+x^4)\,dy + x(1+4y^2)\,dx = 0,\quad y(1)=0 \).
Solution
STEP 1 — Separate
\(\dfrac{dy}{1+4y^2} = -\dfrac{x\,dx}{1+x^4}\).
STEP 2 — Left integral
\(\int\dfrac{dy}{1+(2y)^2} = \tfrac12\arctan(2y)\) — the \(\tfrac12\) comes from \(d(2y)=2\,dy\).
\(2y = \tan\!\big(\tfrac{\pi}{4} - \arctan x^2\big)\). Using \(\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}\) with \(\tan A = 1\), \(\tan B = x^2\):
\(2y = \dfrac{1-x^2}{1+x^2}\), so
\(\boxed{\;y = \dfrac{1-x^2}{2(1+x^2)}\;}\)
Check
\(y(1) = 0\) ✓. Differentiating gives \(y' = \dfrac{-2x}{(1+x^2)^2}\), and \(1+4y^2 = \dfrac{2(1+x^4)}{(1+x^2)^2}\); substituting, \((1+x^4)y' + x(1+4y^2) = 0\) ✓. Valid for all real \(x\).
6
Escape velocity. By universal gravitation, \(F = -\dfrac{mgR^2}{(x+R)^2}\) where \(x(t)\) is height above the surface, \(R\) the earth's radius, \(g\) surface gravity; and \(F = ma = m\dfrac{dv}{dt}\), so \(m\dfrac{dv}{dt} = -\dfrac{mgR^2}{(x+R)^2}\). (a) Fired upward with initial velocity \(v_0\) and reaching maximum height \(h\), show \(v_0 = \sqrt{\dfrac{2gRh}{R+h}}\). (Hint: chain rule, \(m\frac{dv}{dt} = mv\frac{dv}{dx}\).)(b) Compute \(v_e = \lim_{h\to\infty} v_0\). (c) With \(R = 3960\) mi and \(g=32\) ft/s², find \(v_e\) in ft/s and mi/s.
Solution
STEP 1 — (a) Change the independent variable
As written, the equation mixes \(v(t)\) with \(x(t)\) and cannot be separated. The hint fixes that: \(\dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}\). Substituting and cancelling \(m\):
In feet per second: \(6.9282\times5280 \approx \mathbf{36{,}581\ \text{ft/s}}\). (Equivalently \(\sqrt{2(32)(20{,}908{,}800)}\), since \(3960\ \text{mi} = 20{,}908{,}800\ \text{ft}\).)
Sanity check
The textbook figure for earth's escape velocity is about 25,000 mi/hr \(\approx 6.9\) mi/s ✓.
⌕Source & references
Prof. Williams — Lecture Notes §2.2 Separable Variables (both worked examples above).
Zill — §2.2 and its exercise set; the escape-velocity problem is the standard §2.2 application.
Lesson L3 of the 3-week track drills separation on simpler equations first.
Math 275 · §2.3 Prof. Williams
Linear Equations
One method solves every problem in this section. Get the equation into standard form, build the factor that turns the left side into a single derivative, and integrate. The difficulty is never the calculus — it's spotting the standard form.
①The method, once
Start from any equation of the form \( f(x)\dfrac{dy}{dx} + g(x)y = h(x) \).
The five steps
1. Divide by \(f(x)\) to make the coefficient of \(y'\) equal to 1:
\[ \frac{dy}{dx} + P(x)\,y = Q(x),\qquad P=\frac{g}{f},\; Q=\frac{h}{f} \]
2. Multiply by the integrating factor \( I(x)=e^{\int P(x)dx} \):
\[ e^{\int P dx}\frac{dy}{dx} + e^{\int P dx}P(x)\,y = e^{\int P dx}Q(x) \]
3. Undo the product rule — by construction the left side is a derivative:
\[ \frac{d}{dx}\Big[e^{\int P dx}\,y\Big] = e^{\int P dx}Q(x) \]
4. Integrate both sides: \( e^{\int P dx}y = \displaystyle\int e^{\int P dx}Q(x)\,dx + c \) 5. Solve for \(y\):
\[ y(x) = e^{-\int P dx}\left[\int e^{\int P dx}Q(x)\,dx + c\right] \]
Three habits that prevent most lost marks
Skip the constant inside \(I(x)\) — it cancels. Verify the left side really is \((Iy)'\) with one line of product rule before you spend an integral on it. Divide the constant too — \(c\) enters before you divide by \(I\), so it comes out multiplied by \(1/I\), never sitting alone at the end.
②Two things to check before you commit
Is it linear in the other variable?
An equation that is hopeless in \(y\) may be linear in \(x\). Before reaching for anything harder, try writing \(\dfrac{dx}{dy} + P(y)x = f(y)\). Classwork problem 2 is exactly this.
Are terms with \(y\) hiding on the right?
Collect every \(y\) onto the left before identifying \(P\). Classwork problem 3 puts one there on purpose.
③Vocabulary you will be asked for
A transient term is any term of the solution that tends to 0 as \(x\to\infty\). It is the part of the answer that stops mattering; what is left is the steady state.
The interval of existence \(I\) is the largest interval on which \(P\) and \(Q\) are continuous and which contains the initial \(x\). It comes from the standard form, not the original equation — dividing is what creates the excluded points.
Consider \( xy' + (1+x)y = e^{-x}\sin(2x) \). (a) Solve it. (b) What is the largest interval \(I\) of existence for the general solution? (c) Identify any transient terms.
Solution
STEP 1 — Standard form
Divide by \(x\) — that division is the whole reason part (b) has an answer.
Split \(P\) into \(\tfrac1x\) plus \(1\) — do not try to integrate \((1+x)/x\) as one lump. \(\int P\,dx = \ln|x| + x\), so \(\mu = e^{\ln|x|+x} = xe^{x}\).
STEP 3 — Multiply through and watch it collapse
\((xe^{x}y)' = xe^{x}\cdot\dfrac{e^{-x}\sin 2x}{x} = \sin 2x\). Both the \(x\) and the \(e^{x}\) cancel against the right side — that is the payoff.
Read the standard form, not the original. \(P\) and \(Q\) are continuous everywhere except \(x=0\), so the candidates are \((-\infty,0)\) and \((0,\infty)\). With no initial condition pinning a side, take \(I = (0,\infty)\).
STEP 7 — (c) Transient terms
Ask of each term: does it vanish as \(x\to\infty\)? Both do — the factor \(e^{-x}/x\) crushes each one, and \(\cos 2x\) stays bounded so it cannot rescue the second. Every term is transient; the steady state is zero.
Easy to miss
Part (b) is about the divided equation. The original \(xy'+(1+x)y=\dots\) looks perfectly well behaved at \(x=0\); the singular point only appears once you divide to reach standard form.
2
Solve: \( y\,dx - (ye^{y} - 2x)\,dy = 0 \).
Solution
STEP 1 — Try it in \(y\) and fail on purpose
\(\dfrac{dy}{dx} = \dfrac{y}{ye^{y}-2x}\) is neither separable nor linear in \(y\) — the \(ye^{y}\) in the denominator kills both. That is the signal to flip the roles.
STEP 2 — Flip to \(x(y)\)
\(y\dfrac{dx}{dy} = ye^{y} - 2x \;\Longrightarrow\; \dfrac{dx}{dy} + \dfrac{2}{y}x = e^{y}\). Textbook linear, with \(x\) as the dependent variable: \(P(y)=2/y\), \(Q(y)=e^{y}\).
\(\int y^2e^{y}dy = y^2e^{y} - 2\int ye^{y}dy = e^{y}(y^2 - 2y + 2)\). The pattern for \(y^ne^{y}\): differentiate the polynomial down to nothing, alternating signs — here \(y^2-2y+2\).
Substituting back into \(y\,x'(y) = ye^{y}-2x\) returns \(0\) identically ✓. Leaving the answer as \(x\) in terms of \(y\) is correct — do not try to invert it. The differential form \(M\,dx + N\,dy = 0\) gives you two equations for free, \(dy/dx\) and \(dx/dy\); always test both for linearity before deciding a problem is hard.
There is a \(P\) sitting on the right. Move it left before identifying the coefficient — this is the only real trap in the problem.
\(\dfrac{dP}{dt} + (2t-1)P = 4t-2\).
STEP 2 — Integrating factor
\(\int(2t-1)dt = t^2-t\), so \(\mu = e^{t^2-t}\).
STEP 3 — Spot the substitution
\((\mu P)' = (4t-2)e^{t^2-t} = 2(2t-1)e^{t^2-t}\) — the right side is exactly twice the derivative of the exponent, so no integration by parts is needed.
STEP 4 — Integrate
With \(u = t^2-t,\; du = (2t-1)dt\): \(\int 2e^{u}du = 2e^{u} = 2e^{t^2-t} + C\).
STEP 5 — Divide by \(\mu\)
\(P = 2 + Ce^{-(t^2-t)} = 2 + Ce^{t-t^2}\).
STEP 6 — Apply \(P(0)=1\)
\(1 = 2 + Ce^{0} \Rightarrow C = -1\).
\(\boxed{\;P = 2 - e^{t-t^2}\;}\)
Interval & behaviour
Defined for all real \(t\) — nothing was divided by, so there is no singular point. The exponent \(t-t^2\to-\infty\) in both directions, so \(P\to2\) either way: that term is transient and \(2\) is the steady state.
Easy to miss
Reading \(P(t)=2t\) off the left side without first moving the \(P\) across gives \(\mu=e^{t^2}\) and an integral that does not close in elementary functions. If the integration suddenly looks impossible, suspect the standard form.
4
Solve Newton's law of cooling in general constants: \( \dfrac{dT}{dt} = k(T-T_m),\quad T(0)=T_0 \), where \(k, T_m, T_0 \in \mathbb{R}\).
Solution
STEP 1 — Standard form
Expand and move the \(T\) term left. \(k\) and \(T_m\) are constants, so the right side is just a number:
\(T_0 = T_m + C \Rightarrow C = T_0-T_m\) — the initial gap between the object and its surroundings, which is what makes the answer readable.
\(\boxed{\;T(t) = T_m + (T_0-T_m)e^{kt}\;}\)
STEP 6 — Interpret
Physically \(k<0\), so \(e^{kt}\to0\) and \(T\to T_m\): the initial gap decays exponentially and the object settles at ambient temperature, approaching from above if it started hot and from below if it started cold. \(T_m\) is the steady state; \((T_0-T_m)e^{kt}\) is the transient.
Notes
This one is also separable — \(dT/(T-T_m) = k\,dt\) gives the same answer; the linear route avoids the absolute-value bookkeeping the log introduces. If \(T_0=T_m\) the constant is zero and \(T\equiv T_m\), the equilibrium solution, already contained in the general answer. And the sign of \(k\) is doing real work: with \(k>0\) the same formula describes runaway heating.
\(P(x)=2x\) on the whole domain — only the forcing switches. So the integrating factor is the same throughout and you solve the same equation twice with different right-hand sides: \(\mu = e^{\int 2x\,dx} = e^{x^2}\), giving \((e^{x^2}y)' = e^{x^2}f(x)\).
Continuity is the gluing condition, not differentiability — and you cannot have both. The left and right derivatives at \(x=1\) are \(-1.1036\) and \(-2.1036\): they differ by exactly \(1\), the size of the jump in \(f\). The equation forces that, since \(y' = f - 2xy\) and only \(f\) jumps.
5. \(\mu = e^{x^2}\); \(\tfrac12+\tfrac32e^{-x^2}\) on \([0,1)\), then \(\big(\tfrac12e+\tfrac32\big)e^{-x^2}\).
What separates full credit from most of it
Standard form first, every time — problems 1, 3, and 4 all hand you the equation in a disguised arrangement, and \(P\) is wrong until you fix it. Check that the left side really is \((\mu y)'\) — one line of product rule catches a bad integrating factor before you waste an integral on it. Divide the constant too. The interval comes from the divided equation — singular points are created by the division, so they are invisible in the original form. Transient means "\(\to0\) as \(x\to\infty\)" — test each term separately; a bounded oscillation like \(\cos2x\) does not stop a decaying factor from winning.
⌕Source & references
Prof. Williams — Ch. 2.3 Linear Equations, Write-Up (the five-step derivation and the worked example above); Lecture Notes §2.3.
Zill — §2.3, including the piecewise-forcing problem.
Lesson L4 of the 3-week track covers integrating factors with additional practice.
Math 275 · §2.4 Prof. Williams
Exact Equations
Some equations are already the total differential of a hidden function \(f(x,y)\). Recognise that and the "solving" is just undoing a pair of partial derivatives — and when the recognition fails, an integrating factor can sometimes make it work anyway.
①Where the form comes from
An exact equation is a DE we can express as a differential. Start from a surface \(z = f(x,y)\). Differentiating gives
Naming \(\dfrac{\partial f}{\partial x} = M(x,y)\) and \(\dfrac{\partial f}{\partial y} = N(x,y)\) puts it in the standard shape:
Exact equation
\[ M(x,y)\,dx + N(x,y)\,dy = 0 \]
and the solution is simply the level curve it came from:
Its solution
\[ f(x,y) = c \]
②The test
Mixed partials commute — \(f_{xy}=f_{yx}\) whenever both are continuous on the same interval. Since \(M=f_x\) and \(N=f_y\), that gives the necessary condition:
1. Integrate \(M\) with respect to \(x\), treating \(y\) as a constant. The "constant" of integration is an unknown function \(g(y)\). 2. Integrate \(N\) with respect to \(y\), treating \(x\) as a constant. 3. Compare the two results and read off every distinct term. Assemble \(f(x,y)\), set it equal to \(c\), and apply any initial condition. 4. Verify by checking \(f_x = M\) and \(f_y = N\).
⌖ Worked example 1 — an exact IVP (from the write-up)
For non-exact equations we can sometimes find a factor \(\mu(x,y)\) that makes them exact. Two cases are testable by hand:
Two special integrating factors
If \(\dfrac{M_y-N_x}{N}\) is a function of \(x\) only: \[ \mu(x) = e^{\int \frac{M_y-N_x}{N}\,dx} \]
If \(\dfrac{N_x-M_y}{M}\) is a function of \(y\) only: \[ \mu(y) = e^{\int \frac{N_x-M_y}{M}\,dy} \]
Order of operations
Compute both quotients before choosing. Whichever one simplifies to a single-variable expression is the one that works; if neither does, this method does not apply and you need §2.5 substitutions instead.
⌖ Worked example 2 — non-exact (from the write-up)
Solve \( (3xy+y^2) + (x^2+xy)y' = 0 \).
STEP 1 — Differential form and test
\((3xy+y^2)dx + (x^2+xy)dy = 0\). \(M_y = 3x+2y\), \(N_x = 2x+y\). Not equal → not exact.
STEP 2 — Try both quotients
\(\dfrac{N_x-M_y}{M} = \dfrac{2x+y-3x-2y}{3xy+y^2} = \dfrac{-(x+y)}{y(3x+y)}\) — still has both variables, no good.
\(\dfrac{M_y-N_x}{N} = \dfrac{3x+2y-2x-y}{x^2+xy} = \dfrac{x+y}{x(x+y)} = \dfrac1x\) — a function of \(x\) alone ✓.
STEP 3 — Build \(\mu\)
\(\mu(x) = e^{\int \frac1x dx} = e^{\ln x} = x\).
STEP 4 — Multiply through and re-test
\((3x^2y + xy^2)dx + (x^3+x^2y)dy = 0\). Now \(M_y = 3x^2+2xy\) and \(N_x = 3x^2+2xy\) — equal, so it is exact.
The course folder has the §2.4 write-up but no §2.4 classwork sheet, so this set is built to drill the same three skills the write-up covers: the exactness test, reconstructing \(f\), and finding a one-variable integrating factor. Problem 6 is §2.3 classwork problem 2 solved a second way.
1
Test and solve: \( (2x-1)\,dx + (3y+7)\,dy = 0 \).
Solution
STEP 1 — Identify \(M\) and \(N\)
\(M = 2x-1\), \(N = 3y+7\).
STEP 2 — Test
\(M_y = 0\) (no \(y\) in \(M\)) and \(N_x = 0\) (no \(x\) in \(N\)). Equal → exact. Any equation whose \(M\) is \(x\)-only and \(N\) is \(y\)-only is automatically exact — and also separable.
Steps 3–4 ("differentiate and match") and the write-up's "integrate both, compare" give the same answer. Use whichever integral is easier — here \(\int N\,dy\) is just as quick.
3
Test and solve: \( (2y^2x-3)\,dx + (2yx^2+4)\,dy = 0 \).
Solution
STEP 1 — Test
\(M_y = 4yx\) and \(N_x = 4yx\). Exact.
STEP 2 — Integrate \(M\) in \(x\)
\(f = \int(2y^2x-3)dx = x^2y^2 - 3x + g(y)\) — remember \(y\) is a constant here, so \(2y^2\!\int x\,dx = x^2y^2\).
\(f = \int y^2dx = xy^2 + g(y)\); then \(f_y = 2xy + g'(y) = 2xy - y^2e^{y}\), so \(g'(y) = -y^2e^{y}\) and \(g(y) = -e^{y}(y^2-2y+2)\).
\(xy^2 - e^{y}(y^2-2y+2) = c\)
STEP 6 — Compare with §2.3
Solving for \(x\): \(x = \dfrac{e^{y}(y^2-2y+2)+c}{y^2}\) — identical to the answer the integrating-factor-for-linear method gave in §2.3. Notice that the two \(\mu\)'s agree too: the linear route used \(\mu = y^2\) on \(dx/dy + (2/y)x = e^{y}\), which is this \(\mu = y\) times the \(y\) already multiplying \(dx\).
⌕Source & references
Prof. Williams — Ch. 2.4 Exact Equations, Write-Up (the derivation, the test, both integrating-factor formulas, and both worked examples above).
Zill — §2.4 and its exercise set.
Lesson L5 of the 3-week track covers exact equations alongside substitution methods.
Next in the course: §2.5 Solutions by Substitution (homogeneous and Bernoulli equations) and §2.6 A Numerical Method (Euler's method).
Math 275 · Checkpoint Quiz
Quiz — Math 275, §1.1–2.4
Ten questions drawn from the classwork sets and write-ups of the first two chapters. Tap an answer to lock it in and see why. Aim for 8/10 before the exam.
Question 1
Classify \( x^2y''' - e^{y}y' + 3y = 0 \).
Why
Highest derivative \(y'''\) → 3rd order. The dependent variable sits inside a function in \(e^{y}\) (and multiplies \(y'\)) → non-linear. Right side is \(0\) → homogeneous. This is §1.1 classwork #2.
Question 2
Which of these is linear?
Why
In A every appearance of \(y\) and its derivatives is to the first power; \(e^{x}\) is only a coefficient, and coefficients may involve the independent variable however they like. B has \(e^{y}\), C has the product \(y\,y'\), D has \(y^2\) — all non-linear.
Question 3
Substituting \(y=x^{m}\) into \( 2x^2y'' + 3xy' - 15y = 0 \) forces \(m\) to satisfy:
Why
\(y'=mx^{m-1}\), \(y''=m(m-1)x^{m-2}\), and every term collapses to \(x^{m}\): \(2m(m-1)+3m-15 = 2m^2-2m+3m-15 = 2m^2+m-15\). Factoring \((2m-5)(m+3)\) gives \(m=\tfrac52,-3\). The usual slip is forgetting the \(-2m\) from \(m(m-1)\).
Question 4
For \( 2xy' = \sqrt{xy^2-9} \) at the point \((1,3)\), the existence-and-uniqueness theorem:
Why
\(f = \frac{\sqrt{xy^2-9}}{2x}\) and \(f_y = \frac{y}{2\sqrt{xy^2-9}}\). At \((1,3)\) the radicand is \(0\), so \(f_y\) blows up — and no rectangle around the point stays inside the region where \(f\) is even real. The theorem is a sufficient condition, so its failure leaves the question open; D overstates it.
Question 5
In the virga (evaporating raindrop) model, mass loss proportional to surface area gives \(r(t)=kt+r_0\). Newton's law in momentum form, \(F=\frac{d}{dt}(mv)\), then yields:
Why
\(mv' + vm' = mg\) with \(m = k\tfrac43\pi r^3\) and \(m' = k\,4\pi r^2 r'\). Dividing by \(m\) gives \(v' + \frac{3}{r}v = g\); substituting \(r=kt+r_0\) and \(r'=k\) gives A. Choice C is what you would get if the mass were constant — the whole point is that it isn't.
Question 6
Classify the critical points of \( \dfrac{dy}{dx} = y - y^2 + 2 \).
Why
Factor with the minus sign pulled out: \(-(y-2)(y+1)\), so the parabola opens downward and \(f>0\) only between the roots. Arrows flee \(y=-1\) and approach \(y=2\). Derivative test: \(f'(y)=1-2y\) gives \(f'(-1)=3>0\) (repeller), \(f'(2)=-3<0\) (attractor). Two distinct roots rule out semi-stable.
Question 7
Separating \( dy + (y-3)^2\,dx = 0 \) gives \( y = 3 + \dfrac{1}{x-c} \). What else belongs in a complete answer?
Why
Separating required dividing by \((y-3)^2\), which discards the root \(y=3\). That constant function does satisfy the equation (\(dy=0\) and \((y-3)^2=0\)) but no finite \(c\) produces it from the family — the definition of a singular solution.
Question 8
The integrating factor for \( xy' + (1+x)y = e^{-x}\sin(2x) \) is:
Why
Standard form first: \(y' + \frac{1+x}{x}y = \frac{e^{-x}\sin2x}{x}\), so \(P = \frac1x + 1\) and \(\int P\,dx = \ln|x| + x\), giving \(\mu = e^{\ln|x|+x} = xe^{x}\). Split \(P\) into the two pieces — integrating \((1+x)/x\) as one lump is where A and B come from.
Question 9
In \( y = \dfrac{Ce^{-x}}{x} - \dfrac{e^{-x}\cos(2x)}{2x} \), which terms are transient?
Why
Transient means "\(\to0\) as \(x\to\infty\)." The factor \(e^{-x}/x\) drives both terms to zero, and \(\cos2x\) is bounded so it cannot rescue the second. Every term is transient and the steady state is zero. C is the classic misread: a bounded oscillation does not stop a decaying factor from winning.
Question 10
\( (3xy+y^2)\,dx + (x^2+xy)\,dy = 0 \) is not exact. Which integrating factor fixes it?
Why
\(M_y = 3x+2y\), \(N_x = 2x+y\). Test the \(x\)-quotient: \(\frac{M_y-N_x}{N} = \frac{x+y}{x(x+y)} = \frac1x\), a function of \(x\) alone, so \(\mu = e^{\int dx/x} = x\). Multiplying gives \((3x^2y+xy^2)dx + (x^3+x^2y)dy = 0\) with \(M_y = N_x = 3x^2+2xy\) ✓, and the solution \(x^3y + \tfrac12x^2y^2 = c\).