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.
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.
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).