Ordinary Differential Equations

Initial-value problems, first-order methods, linear higher-order equations, forcing and resonance, and linear systems, with explicit checks for lost solutions, singular models, and uniqueness.

An ordinary differential equation constrains an unknown function of one independent variable and its derivatives. A formula is not a complete solution until its interval, regularity, initial data, and any excluded or singular solutions are accounted for.

A rate law turns a mechanism into a prediction #

An ODE says how the present state controls its instantaneous rate of change. The initial condition identifies the state now; solving the equation converts that local rate rule into a trajectory through time.

The cooling model assumes one uniform object temperature, constant room temperature, and a constant lumped cooling-rate coefficient. The same pattern—state, rate law, initial state, trajectory—appears in motion, circuits, mixing, population models, and feedback control, but every application brings its own assumptions.

Equations, solutions, and initial data #

A first-order equation in normal form is

y′=f(t,y).y'=f(t,y).

An initial-value problem (IVP) adds y(t0)=y0y(t_0)=y_0.

A local existence-and-uniqueness theorem is a contract, not decoration. A common version says that if ff is continuous near (t0,y0)(t_0,y_0) and locally Lipschitz in yy there—for example, if fyf_y is continuous—then a unique local solution passes through the initial point.

If continuity or the Lipschitz condition fails, three different outcomes are possible:

  • no solution;
  • one solution;
  • multiple solutions.

The theorem gives sufficient conditions. Failure of its hypotheses does not, by itself, decide which outcome occurs.

For a normalized linear second-order IVP

y′′+p(t)y′+q(t)y=g(t),y(t0)=a,y′(t0)=b,y''+p(t)y'+q(t)y=g(t),\qquad y(t_0)=a,\quad y'(t_0)=b,

continuity of p,q,gp,q,g on an interval containing t0t_0 gives a unique solution on that interval.

Separable equations and lost solutions #

If

y′=g(t)h(y),y'=g(t)h(y),

then, where h(y)≠0h(y)\ne0,

dyh(y)=g(t) dt.\frac{dy}{h(y)}=g(t)\,dt.

The condition h(y)≠0h(y)\ne0 is load-bearing. Every root h(c)=0h(c)=0 must be checked separately, because y(t)≡cy(t)\equiv c may be an equilibrium solution discarded by division.

A robust workflow is:

  1. list equilibrium solutions h(y)=0h(y)=0;
  2. separate only on regions where h(y)≠0h(y)\ne0;
  3. integrate and keep absolute values where required;
  4. impose initial data;
  5. state the maximal interval before the solution reaches a singularity.

For the logistic equation y′=y(1−y)y'=y(1-y), the sign of the right-hand side sends positive trajectories toward y=1y=1, while y=0y=0 and y=1y=1 are themselves constant solutions. They must be recorded before division by y(1−y)y(1-y).

Solution flow and equilibrium solutions of the logistic equation For y prime equals y times one minus y, the phase line points toward the stable equilibrium y equals one and away from y equals zero on the positive interval. Solution curves do not cross either equilibrium. Dividing by y times one minus y therefore requires recording both constant solutions first. Phase line Noncrossing solution curves y = 0 y = 1 y′ < 0 y′ > 0 y′ < 0 y′ = y(1 − y) t y y ≡ 1 y ≡ 0 t₀ 0 < y₀ < 1 flows upward y₀ > 1 flows downward Unique solution curves cannot cross either equilibrium. Before dividing by y(1 − y), record y ≡ 0 and y ≡ 1; the division assumes y(1 − y) ≠ 0.
Equilibrium solutions are part of the solution family, and uniqueness prevents neighboring trajectories from crossing them. Dividing by the equilibrium factor would silently discard both constant solutions.

Nonuniqueness at a singular slope #

Consider the equation

y′=−1−y21−x2,y(0)=1.y'=-\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}, \qquad y(0)=1.

The constant function y≡1y\equiv1 satisfies the IVP. Separating variables divides by 1−y2\sqrt{1-y^2}, so it cannot recover that solution. Moreover the right-hand side is not locally Lipschitz in yy at y=1y=1, and branch/domain restrictions matter. A nonconstant branch cannot be reported as the unique IVP solution.

Exact equations #

A differential form

M(x,y) dx+N(x,y) dy=0M(x,y)\,dx+N(x,y)\,dy=0

is exact on a suitable domain if there is a potential Φ\Phi with

Φx=M,Φy=N.\Phi_x=M,\qquad \Phi_y=N.

Then the implicit solution is

Φ(x,y)=C.\Phi(x,y)=C.

On a simply connected domain with continuous first partial derivatives, My=NxM_y=N_x is sufficient for exactness. On a domain with holes, equal cross-partials need not yield a global single-valued potential.

To construct Φ\Phi, integrate MM with respect to xx:

Φ(x,y)=∫M(x,y) dx+g(y),\Phi(x,y)=\int M(x,y)\,dx+g(y),

then differentiate with respect to yy and solve for g′(y)g'(y) using NN.

First-order linear equations #

For

y′+p(x)y=q(x),y'+p(x)y=q(x),

choose the integrating factor

μ(x)=e∫p(x) dx.\mu(x)=e^{\int p(x)\,dx}.

Because μ′=μp\mu'=\mu p,

(μy)′=μy′+μp y=μq.(\mu y)'=\mu y'+\mu p\,y=\mu q.

Hence

y(x)=1μ(x)(C+∫μ(x)q(x) dx).y(x)=\frac{1}{\mu(x)} \left(C+\int \mu(x)q(x)\,dx\right).

The method works on intervals where pp and qq are continuous. An integrating factor may have different expressions on disconnected intervals.

Worked example #

Solve

y′+2xy=e−x2xsin⁡x.y'+2xy=e^{-x^2}x\sin x.

The integrating factor is

μ=e∫2x dx=ex2.\mu=e^{\int2x\,dx}=e^{x^2}.

Therefore

(ex2y)′=xsin⁡x.(e^{x^2}y)'=x\sin x.

Integration by parts gives

∫xsin⁡x dx=sin⁡x−xcos⁡x,\int x\sin x\,dx=\sin x-x\cos x,

so

y=e−x2(C+sin⁡x−xcos⁡x).\boxed{y=e^{-x^2}\left(C+\sin x-x\cos x\right)}.

Substitution differentiates the product and recovers the original equation, which is the decisive check on the integrating-factor calculation.

Bernoulli equations #

A Bernoulli equation is

y′+p(x)y=q(x)yn,n≠0,1.y'+p(x)y=q(x)y^n,\qquad n\ne0,1.

On an interval where y≠0y\ne0, set v=y1−nv=y^{1-n}. Since

v′=(1−n)y−ny′,v'=(1-n)y^{-n}y',

multiplying the ODE by (1−n)y−n(1-n)y^{-n} gives a linear equation:

v′+(1−n)p(x)v=(1−n)q(x).v'+(1-n)p(x)v=(1-n)q(x).

Check y=0y=0 separately when the original equation makes sense there. Describing y=0y=0 as the case “C=∞C=\infty” is not legitimate: infinity is not a value of the integration constant in the derived solution family.

Higher-order linear equations #

A linear equation has the form

an(t)y(n)+⋯+a1(t)y′+a0(t)y=g(t),an(t)≠0.a_n(t)y^{(n)}+\cdots+a_1(t)y'+a_0(t)y=g(t), \qquad a_n(t)\ne0.

After division by ana_n, existence and uniqueness are controlled on intervals where all normalized coefficients and gg are continuous.

For the homogeneous constant-coefficient equation

any(n)+⋯+a1y′+a0y=0,a_ny^{(n)}+\cdots+a_1y'+a_0y=0,

try y=erty=e^{rt}. This produces the characteristic polynomial

anrn+⋯+a1r+a0=0.a_nr^n+\cdots+a_1r+a_0=0.

The solution contributions are:

  • real root rr of multiplicity mm:

    ert, tert, …,tm−1ert;e^{rt},\,te^{rt},\,\ldots,t^{m-1}e^{rt};

  • conjugate roots α±iβ\alpha\pm i\beta, each of multiplicity mm:

    tkeαtcos⁡βt,tkeαtsin⁡βt,0≤k<m.t^k e^{\alpha t}\cos\beta t,\quad t^k e^{\alpha t}\sin\beta t,\quad 0\le k<m.

The number of linearly independent basis functions must equal the order of the equation.

Nonhomogeneous equations and superposition #

For a linear operator LL,

L[y]=g,L[y]=g,

the general solution is

y=yh+yp,y=y_h+y_p,

where L[yh]=0L[y_h]=0 and L[yp]=gL[y_p]=g.

This follows because

L[yh+yp]=L[yh]+L[yp]=g.L[y_h+y_p]=L[y_h]+L[y_p]=g.

The forcing gg is not itself a solution; confusing the input with the response is a category error.

Variation of parameters finds ypy_p for general forcing. Undetermined coefficients is faster for polynomial, exponential, sine, and cosine inputs, but only after accounting for resonance.

If the naive trial duplicates a homogeneous mode, multiply the trial by enough powers of tt to make it linearly independent. Resonance is therefore a linear-dependence problem, not a mysterious extra rule.

For an ideal undamped oscillator driven at its natural frequency, the resonant particular solution grows proportionally to tt. Real damping changes the response and must not be omitted when making physical claims.

Linear systems #

A constant system is

x′=Ax.x'=Ax.

If Av=λvAv=\lambda v, then

x(t)=eλtvx(t)=e^{\lambda t}v

is a solution. Distinct eigenvectors generate independent modes.

A repeated eigenvalue does not automatically provide enough eigenvectors. If AA is defective, find a generalized eigenvector ww satisfying

(A−λI)w=v.(A-\lambda I)w=v.

A Jordan-chain pair of solutions is then

eλtv,eλt(tv+w).e^{\lambda t}v, \qquad e^{\lambda t}(tv+w).

Writing four arbitrary constants for a two-dimensional repeated-root system is impossible: a first-order 2×22\times2 system has a two-dimensional solution space.

For nonhomogeneous systems

x′=A(t)x+g(t),x'=A(t)x+g(t),

a fundamental matrix Φ(t)\Phi(t) gives variation of constants:

x(t)=Φ(t)(c+∫Φ(t)−1g(t) dt).x(t)=\Phi(t) \left(c+\int \Phi(t)^{-1}g(t)\,dt\right).

Why eigenvectors matter: two coupled rooms #

Consider two thermal zones with equal effective heat capacities that exchange heat but, in this simplified model, lose no heat outdoors:

T1′=−k(T1−T2),T2′=k(T1−T2).T_1'=-k(T_1-T_2), \qquad T_2'=k(T_1-T_2).

In matrix form,

[T1′T2′]=[−kkk−k][T1T2].\begin{bmatrix}T_1'\\T_2'\end{bmatrix} = \begin{bmatrix}-k&k\\k&-k\end{bmatrix} \begin{bmatrix}T_1\\T_2\end{bmatrix}.

The eigenvector (1,1)(1,1) has eigenvalue 00: equal temperatures do not change, and the average is conserved. The eigenvector (1,−1)(1,-1) has eigenvalue −2k-2k: a temperature difference decays exponentially.

If k=0.1 min−1k=0.1\ \mathrm{min^{-1}}, T1(0)=30∘CT_1(0)=30^\circ\mathrm C, and T2(0)=20∘CT_2(0)=20^\circ\mathrm C, then

T1(t)=25+5e−0.2t,T2(t)=25−5e−0.2t.T_1(t)=25+5e^{-0.2t}, \qquad T_2(t)=25-5e^{-0.2t}.

After ten minutes they are approximately 25.68∘C25.68^\circ\mathrm C and 24.32∘C24.32^\circ\mathrm C. Eigenvectors reveal two independent physical patterns—a conserved average and a decaying difference—that are hidden when the coupled equations are read one row at a time.

Models and singular initial conditions #

A differential equation inherits the validity limits of the physical law used to derive it. For example, a force law containing 1/v1/v is undefined at v=0v=0. Algebraically finding a curve that approaches v=0v=0 does not make an IVP posed at v(0)=0v(0)=0 valid. The model must be reformulated or its domain restricted before applying an existence theorem.

Likewise, exponential decay in an RLRL circuit assumes fixed R,LR,L, a specified voltage input, and an ideal lumped model. State these assumptions before interpreting long-time behavior.

Verification checklist #

For every proposed ODE solution:

  1. substitute it into the original equation, not only a transformed equation;
  2. check all initial or boundary data;
  3. identify divisions and substitutions that may have removed solutions;
  4. state the interval and singular points;
  5. confirm that the number of arbitrary constants matches the order;
  6. for systems, verify eigenvectors or generalized eigenvectors directly;
  7. separate mathematical behavior from model assumptions.

Failure modes #

  • Dividing by yy, 1−y21-y^2, or another solution-dependent factor without checking its zeros.
  • Writing a singular solution as a limiting “infinite constant.”
  • Claiming uniqueness when the vector field is not locally Lipschitz at the initial point.
  • Treating vertical curves as solutions of an equation posed for y(x)y(x).
  • Omitting absolute values introduced by logarithmic integration.
  • Using a characteristic-root formula across a point where the leading coefficient vanishes.
  • Forgetting the tt factor for repeated roots or resonance.
  • Treating a repeated eigenvalue as if it automatically supplied a full eigenbasis.
  • Reporting only a particular solution to a nonhomogeneous equation.
  • Applying a physical ODE at an initial state where its own right-hand side is undefined.

Sources

  1. bookDifferential and Integral Calculus of Functions of Several Variables and Differential Equations: Lecture Notes and Practicumpdfnotes
  2. docsMIT 18.03SC Differential Equations Lecture Notes
  3. bookMIT ES.1803 Spring 2024 Topic Notes
  4. docsMIT 18.03SC Unit II: Second-Order Constant-Coefficient Linear Equations
  5. docsMIT 18.03SC Damped Harmonic Oscillators
  6. docsMIT 18.03 Notes and Exercises: Linear Systems
  7. bookOpenStax Calculus Volume 2 — Separable Equations and Newton's Law of Cooling

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