Ten lessons connecting ODE rules, methods, approximation, forcing and systems, each with paired explorations.
Introductory ODE bridge, not a full semester. PDEs, full Fourier series, distribution theory and advanced nonlinear dynamics are outside scope. No jurisdiction-specific certification standard is claimed.
Single-variable derivatives and integrals, exponentials, algebra; basic matrices for the last lesson.
Course outline
A slope is not a solution
Check both the equation and the initial value.
Do not divide away an equilibrium
Separate variables while retaining excluded constant solutions.
The product-rule disguise
Solve a first-order linear equation with an integrating factor.
Arrows before formulas
Classify scalar autonomous equilibria from signs.
Euler takes a tangent-sized step
Compute explicit Euler and separate approximation from exact solution.
A balance writes the equation
Build a hypothetical mixing model with units and assumptions.
Two constants, two starting facts
Solve second-order constant-coefficient IVPs including repeated roots.
Forcing is not the whole response
Add homogeneous and particular terms and check resonant overlap.
Laplace keeps the starting value
Retain initial terms in a one-sided Laplace solve.
Systems move in modes
Solve a simple matrix system and inspect every growth direction.
Sources and curriculum note
Sources fetched October 4, 2026. These are archived courses, not a current universal syllabus. All teaching scenarios are explicitly hypothetical.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. A slope is not a solution
Learning goal: Check both the equation and the initial value.
An ODE prescribes a derivative in terms of the unknown function and independent variable. A solution must obey the rule at every point of its interval. Passing through an initial point does not itself satisfy the equation. Differentiate a candidate, substitute it, and check the initial condition separately.
A direction field assigns slope f(t,y) at each point of y′=f(t,y). Its short segments describe local directions, not entire solution curves. Continuity of f and its partial derivative with respect to y near the initial point is a sufficient condition for local existence and uniqueness. Local is not global: singularities or blow-up can limit the solution interval.
In hypothetical y′=y, y(0)=2, y=2eᵗ passes both tests. Constant y=2 passes the initial test but has zero derivative. The coefficient and starting value are invented teaching choices, not a measured growth process. State the interval whenever a formula could become undefined.
Official source grounding: DE1, DE2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Test hypothetical y=2eᵗ for y′=y, y(0)=2.
Differentiate: y′=2eᵗ.
Substitute y: right side equals 2eᵗ, so the identity holds.
At zero, y=2e⁰=2.
The formula is defined on all real t and solves this IVP there.
Practice problem and solution
New hypothetical y′=-0.5y, y(0)=8. Confirm y=8e^(-0.5t) fits both the equation and the starting value, then enter the time t when y reaches 2, to 4 decimals.
y′=-0.5·8e^(-0.5t)=-0.5y and y(0)=8, so the candidate fits. Setting 8e^(-0.5t)=2 gives e^(-0.5t)=1/4, so t=ln4/0.5=2.7726.
Mental model: A solution obeys a derivative identity on an interval.
Common trap: Checking only the starting point.
2. Do not divide away an equilibrium
Learning goal: Separate variables while retaining excluded constant solutions.
For y′=g(t)h(y), separation gives dy/h(y)=g(t)dt only where h(y) is nonzero. First find zeros of h and check those constant functions in the original equation. Division can remove genuine equilibria. The separated family and the excluded constant solutions together describe the possibilities.
Integrate both sides, retain a constant, and use the initial value. Absolute values in logarithms and signs after exponentiation matter. The solved formula must satisfy the ODE and remain defined on a connected interval containing the initial time. A nearby existence theorem cannot promise a finite answer after blow-up.
For hypothetical y′=y², y(0)=1, integration gives -1/y=t+C and y=1/(1-t). The interval containing zero ends at t=1. Constant y=0 also solves the ODE but not this initial value. No real process is asserted: this is an exact toy example of a solution with a finite singularity.
Official source grounding: DE1, DE2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Solve hypothetical y′=y², y(0)=1.
Record equilibrium y=0 before division.
Integrate y⁻²dy=dt: -1/y=t+C.
Use y(0)=1 to get C=-1, hence y=1/(1-t).
Its derivative equals y², and the maximal interval containing zero is (-∞,1).
Practice problem and solution
New hypothetical y′=y(3-y), y(0)=1. Constant solutions occur at y=0 and y=3. Using y=3/(1+2e^(-3t)), enter y(1) to 4 decimals.
At t=0: 3/(1+2)=1, so the starting value holds. At t=1: e^(-3)=0.049787, so y=3/(1+0.099574)=2.7283. The solution stays between the equilibria 0 and 3.
Mental model: Division can hide solutions; domains can end.
Common trap: Extending past a denominator zero.
3. The product-rule disguise
Learning goal: Solve a first-order linear equation with an integrating factor.
Write the equation as y′+p(t)y=q(t) before choosing μ=exp(∫p dt). Then μ′=pμ, and multiplication turns the left side into (μy)′. Integrate this full product equation, divide by μ, and determine the constant from the initial condition.
If the coefficient of y′ is not one, normalize on an interval where that coefficient is nonzero. Any nonzero constant multiple of μ works. Check μ′=pμ to catch a sign error: a decaying transient in the answer does not imply a negative exponent in the integrating factor.
In hypothetical y′+2y=6, y(0)=1, μ=e²ᵗ yields y=3+Ce⁻²ᵗ. Initial data gives C=-2. The steady value three comes from forcing divided by the coefficient, while the initial offset decays. This is a symbolic toy system, not a claimed physical observation.
Official source grounding: DE1, DE2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Solve hypothetical y′+2y=6, y(0)=1.
μ=e²ᵗ, so (e²ᵗy)′=6e²ᵗ.
Integrate: e²ᵗy=3e²ᵗ+C.
Divide and use y(0)=1: y=3-2e⁻²ᵗ.
y′=4e⁻²ᵗ; adding 2y gives 6 and y(0)=1.
Practice problem and solution
New hypothetical y′+3y=9eᵗ, y(0)=1. Using the integrating factor e^(3t), enter y(1) to 4 decimals.
(e^(3t)y)′=9e^(4t), so e^(3t)y=(9/4)e^(4t)+C and y=(9/4)eᵗ+Ce^(-3t). y(0)=1 gives C=-5/4. Then y(1)=2.25e-1.25e^(-3)=6.1161-0.0622=6.0539.
Mental model: The integrating factor creates a product derivative.
Common trap: Choosing p before normalization.
4. Arrows before formulas
Learning goal: Classify scalar autonomous equilibria from signs.
For y′=f(y), equilibrium y* means f(y*)=0. Between equilibria, positive f gives increasing y and negative f gives decreasing y. Stability asks what nearby trajectories do, not whether the equilibrium number is positive. Test one point in each interval after finding the roots.
In the usual smooth scalar setting, arrows toward an equilibrium from both sides indicate asymptotic stability; arrows away indicate instability. f′(y*)<0 is a sufficient local attraction test and f′(y*)>0 indicates repulsion. If f′=0, the derivative test is inconclusive: inspect signs instead of forcing a conclusion.
The hypothetical dimensionless equation y′=y(1-y) has equilibria zero and one. Signs are negative below zero, positive between zero and one, negative above one. One attracts while zero repels. This toy equation is not a population forecast; a physical interpretation would require a justified domain and parameters.
Official source grounding: DE1, DE2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Classify hypothetical y′=y(1-y).
Roots are zero and one.
At -1, 1/2, 2, f is -2, 1/4, -2.
Arrows point away from zero and toward one.
Zero is unstable and one asymptotically stable in the scalar model.
Practice problem and solution
New hypothetical y′=y(6-y)-5, so y′=-(y-1)(y-5). Find the equilibria, then enter f′(y*) at the stable equilibrium, where f(y)=y(6-y)-5.
f(y)=-(y-1)(y-5), so equilibria are 1 and 5. f′(y)=6-2y. f′(1)=4>0 is unstable and f′(5)=-4<0 is stable. The answer is -4.
Mental model: Stability describes nearby motion.
Common trap: Equating positive height with stability.
5. Euler takes a tangent-sized step
Learning goal: Compute explicit Euler and separate approximation from exact solution.
Forward Euler uses yₙ₊₁=yₙ+h f(tₙ,yₙ), and tₙ₊₁=tₙ+h. The slope is evaluated at the beginning of the current step and held constant across that step. Recompute it after updating. The new value is normally an approximation, not an exact ODE solution.
Under appropriate smoothness and stability assumptions, reducing h improves the approximation on a fixed finite interval. It is not an unconditional rule. For y′=-ky, Euler multiplies by 1-kh; numerical decay requires |1-kh|<1. The exact solution can decay while a poorly chosen step grows.
In hypothetical y′=y, y(0)=1, h=1/2, two steps give 1.5 then 2.25 at t=1. The exact e is about 2.71828. The case comparison is original arithmetic, not measured performance. Compare the current-step slope, the updated state and the exact value as separate quantities.
Official source grounding: DE1. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Two Euler steps for hypothetical y′=y, y(0)=1, h=0.5.
At zero, slope one gives y₁=1.5 at t=0.5.
Slope 1.5 gives y₂=1.5+0.75=2.25 at t=1.
Exact solution is eᵗ, so exact y(1)≈2.71828.
Absolute error is about 0.46828; label the computed value approximate.
Practice problem and solution
New hypothetical y′=2t-y, y(0)=1, with exact solution y=2t-2+3e^(-t). Take three Euler steps of h=0.5 to reach t=1.5, then enter exact minus Euler at t=1.5, to 4 decimals.
Learning goal: Build a hypothetical mixing model with units and assumptions.
A conservation balance writes amount rate as inflow minus outflow plus justified sources. State what the unknown means. If y is solute amount and V is tank volume, concentration is y/V. In a perfectly mixed tank, this is also the outflow concentration. Multiply by volume flow to obtain solute amount per time.
Constant volume requires equal inflow and outflow or another stated mechanism. Otherwise use a volume balance V(t). Units must agree in every term. Stop the model before a tank empties or its assumptions fail. A formula is only as appropriate as those assumptions, even if it can be solved exactly.
In a hypothetical 10 L tank with equal 2 L/min flows and input concentration 3 g/L, input is 6 g/min and output is 0.2y g/min. Therefore y′=6-0.2y. Perfect mixing and fixed rates are toy assumptions, not claims about a real tank. Equilibrium y=30 g has concentration 3 g/L.
Official source grounding: DE0, DE2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Hypothetical 10 L mixing tank, 2 L/min each way, 3 g/L entering, y(0)=5 g. Build the ODE.
Input rate is 2·3=6 g/min.
Output is 2(y/10)=0.2y g/min.
y′=6-0.2y, y(0)=5.
At equilibrium 6-0.2y=0, giving 30 g, or concentration 3 g/L.
Practice problem and solution
New hypothetical 50 L tank, 5 L/min enters at 2 g/L and 5 L/min leaves well mixed, y(0)=10 g. Using y′=10-0.1y, enter the solute amount in grams at t=10 min, to 4 decimals.
Inflow is 5·2=10 g/min and outflow is 5·y/50=0.1y g/min. Equilibrium is 100 g, so y=100+Ce^(-0.1t) and y(0)=10 gives C=-90. At 10 min, y=100-90e^(-1)=66.8909 g.
Mental model: Conservation and units define the ODE.
Common trap: Confusing concentration and amount.
7. Two constants, two starting facts
Learning goal: Solve second-order constant-coefficient IVPs including repeated roots.
Trying eʳᵗ in y″+ay′+by=0 yields r²+ar+b=0. Distinct real roots give two exponentials. A repeated root needs (C₁+C₂t)eʳᵗ, not two identical exponentials. Independent solutions are necessary to span the two-constant family.
Second-order initial data normally specifies y(0) and y′(0). Differentiate the general expression before applying the second condition. Complex roots α±iβ yield eᵅᵗ(C₁cos βt+C₂sin βt). Roots encode modes; constants encode initial conditions.
Hypothetical y″+3y′+2y=0 has roots -1 and -2. With y(0)=1, y′(0)=0, the constants are two and minus one. Thus y=2e⁻ᵗ-e⁻²ᵗ. The sum can satisfy both starting facts even though neither single mode can do so with one weight.
Official source grounding: DE1, DE2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Solve hypothetical y″+3y′+2y=0, y(0)=1, y′(0)=0.
Factor (r+1)(r+2), giving y=C₁e⁻ᵗ+C₂e⁻²ᵗ.
Derivative is -C₁e⁻ᵗ-2C₂e⁻²ᵗ.
C₁+C₂=1, -C₁-2C₂=0 give C₁=2, C₂=-1.
Each mode satisfies the ODE; y(0)=1 and y′(0)=0 verify the IVP.
Practice problem and solution
New hypothetical y″+5y′+6y=0, y(0)=2, y′(0)=-3. Write y=C₁e^(-2t)+C₂e^(-3t), find both constants, then enter y(1) to 4 decimals.
Roots of r²+5r+6 are -2 and -3. y(0): C₁+C₂=2. y′(0): -2C₁-3C₂=-3. Then C₂=-1 and C₁=3. y(1)=3e^(-2)-e^(-3)=0.4060-0.0498=0.3562.
Mental model: Independent modes plus initial data fix the response.
Common trap: Duplicating an exponential for a repeated root.
8. Forcing is not the whole response
Learning goal: Add homogeneous and particular terms and check resonant overlap.
For linear Ly=g, the general solution is yh+yp. A particular solution supplies the forcing, but normally does not satisfy the initial conditions alone. Add the homogeneous family, then use initial data. Superposition combines responses only when the operator and forcing are treated consistently.
Undetermined coefficients chooses a trial suggested by the forcing. If the trial overlaps a homogeneous mode, multiply by enough powers of t to remove that overlap. Undamped forcing at the natural frequency can create a growing-amplitude particular response. Damping changes the conclusion; a driven oscillation is not automatically unbounded resonance.
In hypothetical y″+y=sin t, the plain trial A sin t is annihilated by the left side. Instead yp=-(t/2)cos t works. The full family adds C₁cos t+C₂sin t. This is a toy calculation, not a measured oscillation or an engineering safety prediction.
Official source grounding: DE1, DE2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Find yp for hypothetical y″+y=sin t.
Trial A t cos t avoids the plain sinusoidal overlap.
yp′=A cos t-A t sin t.
yp″=-2A sin t-A t cos t; adding yp leaves -2A sin t.
Match -2A=1: A=-1/2. Add the homogeneous family for a general solution.
Practice problem and solution
New hypothetical y″+y=2cos t, y(0)=0, y′(0)=0. The forcing matches a homogeneous mode, so try yp=t sin t. Enter y(π/2) to 4 decimals.
yp=t sin t gives yp″=2cos t-t sin t, so yp″+yp=2cos t. y(0)=0 gives C₁=0 and y′(0)=0 gives C₂=0. So y=t sin t and y(π/2)=π/2≈1.5708.
Mental model: Full response equals homogeneous plus particular.
Common trap: Using initial conditions only on yp.
9. Laplace keeps the starting value
Learning goal: Retain initial terms in a one-sided Laplace solve.
When smoothness and growth conditions permit, the one-sided Laplace transform changes derivatives into algebra. L{y′}=sY-y(0), and L{y″}=s²Y-sy(0)-y′(0). Initial terms carry the starting state; they are not optional afterthoughts.
Solve for Y, express it using recognizable transform pairs, then invert to time t. The transform variable s is not time. A delay u(t-a)f(t-a), a≥0, gives e⁻ᵃˢF(s). Shifting a whole delayed signal is not the same as replacing every s by s-a.
For hypothetical y′+2y=0, y(0)=3, the algebra is (s+2)Y=3. Inversion gives 3e⁻²ᵗ for t≥0. Omitting the initial term incorrectly yields zero. Verify the inverse expression against the original ODE and starting value rather than treating algebraic partial fractions as the full proof.
Official source grounding: DE1, DE2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Solve hypothetical y′+2y=0, y(0)=3 by Laplace.
Let Y=L{y}, so L{y′}=sY-3.
(s+2)Y=3 gives Y=3/(s+2).
Invert using L{e⁻²ᵗ}=1/(s+2), giving y=3e⁻²ᵗ.
y′+2y=0 and y(0)=3 verify both conditions.
Practice problem and solution
New hypothetical y′+3y=6, y(0)=5. Apply the Laplace transform, solve for Y(s), then enter y(0.5) to 4 decimals.
sY-5+3Y=6/s, so Y=(5s+6)/(s(s+3))=2/s+3/(s+3). Then y=2+3e^(-3t) and y(0.5)=2+3e^(-1.5)=2.6694.
Mental model: Laplace turns initial state into algebraic data.
Common trap: Dropping initial terms.
10. Systems move in modes
Learning goal: Solve a simple matrix system and inspect every growth direction.
For x′=Ax with constant A, Av=λv generates a solution eˡᵃᵐᵇᵈᵃᵗv. With an eigenvector basis, decompose the initial vector into modes and evolve each coefficient. Coordinates are independently scalar only when the matrix actually decouples them or after an appropriate basis change.
All eigenvalues with negative real part make a constant linear system asymptotically stable at the origin. A positive real part gives an unstable direction. Zero real parts need care, and nonlinear linearization can be inconclusive there. A saddle has both growing and decaying directions; one approaching trajectory does not make the equilibrium stable.
In hypothetical A=diag(-1,2), x(0)=(3,1), solution is (3e⁻ᵗ,e²ᵗ). Even though the first component decays, the second grows. Starting exactly on the stable axis removes that growing coefficient, but nearby starts need not stay there. The model is mathematical, not a forecast for a real two-species population.
Official source grounding: DE0, DE1, DE2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Solve hypothetical x′=diag(-1,2)x, x(0)=(3,1).
The equations are x₁′=-x₁ and x₂′=2x₂.
General components are C₁e⁻ᵗ and C₂e²ᵗ.
Initial data sets C₁=3, C₂=1.
Solution (3e⁻ᵗ,e²ᵗ) has a growing mode, so the origin is an unstable saddle.
Practice problem and solution
New hypothetical x′=[[1,2],[2,1]]x, x(0)=(5,1). Eigenvalue 3 has vector (1,1) and eigenvalue -1 has vector (1,-1). Enter x₁(0.5) to 4 decimals.
(5,1)=3(1,1)+2(1,-1). So x(t)=3e^(3t)(1,1)+2e^(-t)(1,-1) and x₁(t)=3e^(3t)+2e^(-t). At t=0.5: 3e^(1.5)+2e^(-0.5)=13.4450+1.2131=14.6581.
Mental model: Eigenvalues set growth; initial values set weights.
Common trap: Inferring whole-system stability from one trajectory.