Differential equations and initial conditions

A rate law describes possible histories

Suppose a quantity y changes at a rate proportional to its current amount: y′=ky. This statement specifies a relation at every time, not a value at one time. Different initial amounts produce different histories, even though they obey the same rate law. For constant k, the functions y(t)=Ce^(kt), including C=0, satisfy it. A measured starting amount selects C; the equation alone does not.

An ordinary differential equation, or ODE, relates an unknown function of one independent variable to one or more of its derivatives. The independent variable is often time t, but it can also be position or another coordinate. A partial differential equation instead involves partial derivatives of a function of several independent variables. The distinction concerns the mathematical unknown and its arguments, not whether the equation looks short or complicated.

An initial-value problem combines a differential equation with data at one independent-variable value, such as y′=f(t,y), y(t₀)=y₀. The data identify a point through which a candidate solution must pass. Whether such a solution exists, whether it is unique and how long it remains valid are separate questions. An attractive graph cannot answer all three.

The goal here is to understand and check an initial-value problem before choosing a solution technique. Lesson 138 develops first-order methods, lesson 139 treats linear oscillation and forcing, and lesson 140 studies systems and phase portraits. This introduction keeps model assumptions, mathematical theorems and numerical approximations distinct so that later calculations have a precise object to solve.

A solution includes a domain and an interval

Let f be defined on a domain D in the (t,y)-plane. A classical solution of y′=f(t,y) on an open interval I is a differentiable function y:I→ℝ whose graph lies in D and whose derivative satisfies the equation at every t in I. When f is continuous along that graph, y′ is continuous as well. For an initial-value problem, t₀ belongs to I and y(t₀)=y₀ must also hold. One-sided physical time restrictions can then be stated separately.

To verify a proposed solution, differentiate it, substitute into the equation, check the initial data and identify the interval on which every expression is defined. Passing through (t₀,y₀) alone is insufficient. So is satisfying the differential equation with the wrong initial value. A residual R(t)=y′(t)−f(t,y(t)) is zero for an exact first-order solution; that algebraic check still needs the domain and initial condition.

For y′=y² and y(0)=1, the function y(t)=1/(1−t) has derivative 1/(1−t)² and satisfies both requirements on (−∞,1). The same formula is defined on (1,∞), but that separate branch is not a solution interval containing the initial time 0. Writing ℝ minus the point 1 as though it were a single connected solution interval hides the obstruction.

An explicit function formula is convenient but not required for understanding a solution. An implicit relation can describe a branch that becomes a differentiable function locally, and some solutions have no convenient elementary closed form. Conversely, an implicit curve can have several branches or vertical tangents. One must identify the actual function and interval rather than assume every point of a plotted relation is one classical solution.

Order, linearity and autonomy answer different questions

The order is the highest derivative order appearing in the differential equation. Thus y′+y=t is first order, while y″+(y′)²=0 is second order. Squaring a first derivative does not make it a second derivative: (y′)²+y=0 is still first order. The order alone does not guarantee a particular number of arbitrary constants or a unique initial-value solution when coefficients are singular or the equation is degenerate.

A scalar linear equation has the form aₙ(t)y^(n)+⋯+a₁(t)y′+a₀(t)y=g(t), with coefficients depending only on the independent variable. Terms such as y², yy′ or sin y make an equation nonlinear in its unknown. The equation y′+t²y=sin t is linear, despite its nonconstant coefficients and trigonometric forcing. A linear equation is homogeneous when g=0 after it has been written in this form.

A first-order equation in normal form explicitly supplies the derivative: y′=f(t,y). A more general relation H(t,y,y′)=0 may not determine one derivative value. For example (y′)²=1 admits both slopes +1 and −1; it is not one single-valued slope field until a branch is selected. Dividing a coefficient to obtain normal form can also exclude points where that coefficient vanishes.

An autonomous normal-form equation is y′=g(y), with no explicit t dependence. Its slope at a state depends only on that state. Nonautonomous equations can depend explicitly on t, as y′=t−y does. Autonomy and linearity are independent classifications: y′=y(1−y) is autonomous and nonlinear, while y′+ty=0 is linear and nonautonomous.

Initial data specify a state, with regularity conditions attached

For a regular scalar first-order normal-form problem, one initial value y(t₀)=y₀ is the usual datum. For a regular nth-order equation solved for its highest derivative, a natural initial state is the list y(t₀),y′(t₀),…,y^(n−1)(t₀). For the associated first-order system x′=F(t,x), continuity and a local Lipschitz bound in the state, ‖F(t,u)−F(t,v)‖≤L‖u−v‖ uniformly in nearby t on a suitable neighbourhood, yield a local unique solution. This is the vector counterpart of the scalar theorem stated later. Merely counting n numbers is not a substitute for those assumptions.

For a particle with position q satisfying q″=−ω²q, its state is position and velocity. Specifying q(0) without q′(0) leaves many possible motions. Setting x₁=q and x₂=q′ rewrites the equation as x₁′=x₂, x₂′=−ω²x₁. This is the bridge to lesson 140: a higher-order scalar equation becomes a first-order system by retaining the needed derivatives as state variables.

Initial conditions are imposed at the same initial time. Conditions at two different times or positions instead form a boundary-value problem, which can have different existence and uniqueness properties. For instance q″+q=0 with q(0)=q(π)=0 has all solutions q=A sin t, so those two endpoint values do not determine A. They are not interchangeable with position and velocity at time 0.

Degeneracy can defeat an informal data count. The relation t y′=2y is satisfied at t=0 only if y(0)=0; it admits many functions y=Ct² with that initial value. Dividing by t yields y′=2y/t only where t≠0, so a normal-form uniqueness theorem near t=0 cannot be invoked by that division. The original equation and its divided version have different stated domains.

Build a model by balancing what enters and leaves

Let a well-mixed tank have fixed liquid volume V, inflow and outflow volume rate q, inflow solute concentration c_in and retained solute mass m(t). Assume perfect mixing, equal inflow and outflow, constant V and no reaction or precipitation. The outgoing concentration is m/V. Over a short time, mass enters at q c_in and leaves at q m/V, giving m′=q c_in−(q/V)m.

If V is measured in litres, q in litres per minute and concentration in grams per litre, both terms on the right have units grams per minute, matching m′. The coefficient q/V has units inverse minutes. This check exposes a common error: subtracting a concentration directly from a mass rate. Initial mass m(t₀)=m₀ is additional information, rather than another term in the rate law.

The equilibrium mass is m_*=V c_in for constant inputs. A candidate solution is m(t)=m_*+(m₀−m_*)e^(−(q/V)(t−t₀)). Differentiate and substitute to verify it, and then check m(t₀)=m₀. For positive V and q and nonnegative concentrations and starting mass, the future solution stays between its starting mass and equilibrium, which supports its physical interpretation under these assumptions.

If the tank volume changes, the outgoing concentration is m/V(t), and V(t) needs its own balance. If mixing is imperfect, a single concentration may not describe the tank. A verified solution of the simplified ODE establishes what that model predicts; it does not establish that the mixing assumptions fit an actual tank. Mathematical correctness and empirical adequacy are different checks.

Cooling distinguishes a state from its difference to equilibrium

For an object in a constant-temperature environment, a simple cooling model is T′=−k(T−T_a), k>0. Here T_a is the ambient temperature and k has units inverse time. The rate is proportional to the temperature difference, not the absolute numerical temperature. This empirical model assumes conditions under which one bulk object temperature and a constant effective coefficient are adequate.

Define the temperature difference θ=T−T_a. Because T_a is constant, θ′=−kθ. With T(t₀)=T₀, the candidate is T(t)=T_a+(T₀−T_a)e^(−k(t−t₀)). Direct differentiation gives T′=−k(T₀−T_a)e^(−k(t−t₀)), exactly −k(T−T_a). The exponential argument is dimensionless, and the initial condition is satisfied.

For future times, a hotter object decreases toward ambient temperature and a colder object increases toward it. In this exact model, a nonzero initial difference never becomes exactly zero at a finite time; it decays exponentially. A measured threshold such as being within one degree is therefore a different question from exact equality. The difference halves after a time ln 2/k, regardless of its starting magnitude.

If ambient temperature varies, θ′=T′−T_a′ rather than simply T′. The previous constant-ambient solution is then not automatically valid. The model also does not provide a universal k for every material, temperature range or transfer mechanism. These limitations should be stated before interpreting a fitted exponential as a complete physical law.

Population assumptions determine the equation and its scale

A continuous population approximation may start with per-capita net growth rate r, giving N′=rN. If that per-capita rate decreases linearly with population and vanishes at N=K>0, one model is N′=rN(1−N/K), with r>0. This logistic equation assumes a well-mixed population, fixed effective conditions and a continuous population variable. It does not explicitly include age structure, discrete individuals, delay, migration or random fluctuations.

N and K have the same population units, r has units inverse time, and 1−N/K is dimensionless. N=0 and N=K are equilibrium solutions. Below K but above zero the right side is positive; above K it is negative. These signs describe the instantaneous rate. They do not themselves specify the population at a future time without initial information and a solution argument.

Introduce u=N/K and τ=r(t−t₀). The chain rule gives du/dτ=u(1−u). This nondimensionalisation separates the scales K and 1/r from the shape of the equation. The initial value is u(0)=N₀/K. A graph using τ and u must label those dimensionless axes; its horizontal coordinate is not automatically measured in days and its vertical coordinate is not the original number of individuals.

K is an equilibrium of this idealised law, not a universal maximum population in every model or every environment. With N₀>K the differential equation is still mathematically meaningful and predicts a decrease toward K in forward time. A fitted curve can be a useful summary while its extrapolation fails when resources, interactions or parameters change. Subsequent lessons can solve the equation, but these modelling qualifications already belong to its formulation.

A slope field gives local directions, not a completed history

For y′=f(t,y), a slope field places a short segment of slope f(t,y) at each sampled point where the field is defined. A solution graph is tangent to these directions at every point on its interval. In a (t,y)-plane, the extended tangent direction can be represented by (1,f(t,y)); it is different from a vector field whose arrows describe velocities in a two-component state space.

For the autonomous equation y′=g(y), an equilibrium y_* satisfies g(y_*)=0, so the constant function y(t)=y_* is a solution. The sign of g above and below an equilibrium indicates whether nearby nonconstant solutions initially increase or decrease. For y′=−2(y−3), slopes are positive below 3 and negative above 3, and exact solutions approach 3 for future times.

Under local uniqueness, two solutions meeting at the same (t,y) agree on their common local interval. Thus distinct solution histories of one such scalar problem cannot cross in the same time-state plane and continue differently. The time coordinate matters: two trajectories can attain the same y value at different times. Without uniqueness, a direction field can admit several histories through one point, so a non-crossing claim requires its hypotheses.

Pictures sample finitely many slopes and can miss singularities, fast variation or delicate long-time behaviour. A plotted equilibrium line is supported by substituting its constant value into g. A stability conclusion requires an argument about nearby solutions and time direction, not only arrows chosen at a few sample points. Lesson 140 extends the geometry to systems, where projected pictures need further care.

The integral equation retains the initial value

If y is a classical solution and f is continuous along its graph, the fundamental theorem of calculus gives y(t)=y₀+∫_(t₀)^t f(s,y(s)) ds. The integrand contains the unknown history y(s). This is not simply an antiderivative of a known function of s, but it expresses the initial-value problem in a useful equivalent form.

Conversely, suppose y is continuous, its graph lies in the field domain, f(s,y(s)) is continuous and the displayed integral equation holds on an interval. The fundamental theorem makes y differentiable and gives y′=f(t,y); at t=t₀ the integral is zero, so the initial value is also recovered. These continuity conditions justify the equivalence used here. More general weak or discontinuous formulations require their own definitions.

The integral form also displays how errors accumulate. A rate difference contributes through an integral over time, while an initial discrepancy enters directly as a starting offset. This motivates both iteration and sensitivity estimates. It does not say that every guessed continuous curve solves the equation: the curve must equal the result of inserting that same curve into the integral.

For y′=y and y(0)=1, start with u₀(t)=1 and define uₙ₊₁(t)=1+∫_0^t uₙ(s) ds. The first iterates are 1+t, 1+t+t²/2 and 1+t+t²/2+t³/6. They suggest the exponential series. The next section states a condition ensuring local convergence, rather than inferring it from the first few polynomial terms.

A local theorem has a quantitative neighbourhood

Take a closed rectangle R=[t₀−a,t₀+a]×[y₀−b,y₀+b], with a,b>0, contained in the field domain. Suppose f is continuous on R and uniformly Lipschitz in y there: |f(t,u)−f(t,v)|≤L|u−v| for all relevant t,u,v, with finite L≥0. Then the initial-value problem has a unique local solution near (t₀,y₀). This is a sufficient local Picard existence and uniqueness theorem, not a necessary test and not a claim of existence for every time.

Continuity on the compact rectangle gives a finite bound |f|≤M. Choose h>0 such that h≤a, Mh≤b and q=Lh<1. On continuous functions u:[t₀−h,t₀+h]→[y₀−b,y₀+b], define (Tu)(t)=y₀+∫_(t₀)^t f(s,u(s)) ds. The bound Mh≤b keeps Tu inside that same vertical interval. In the maximum norm, ‖Tu−Tv‖∞≤q‖u−v‖∞. The integral operator is therefore a contraction on this set.

Starting with u₀(t)=y₀ and iterating uₙ₊₁=Tuₙ, consecutive differences are bounded by qⁿ‖u₁−u₀‖∞. The geometric tail makes the sequence uniformly Cauchy. Completeness of the real numbers gives pointwise limits, and the same tail estimate gives uniform convergence to a continuous function in the closed vertical interval. The Lipschitz estimate allows passage to the limit in the integral equation, so the limit is a fixed point. It is a classical solution on the open interior (t₀−h,t₀+h), with the endpoint values supplied by its continuous extension to the closed interval.

If two fixed points lie in this set, their distance is at most q times itself, so it must be zero. Any continuous solution through the initial point stays in this rectangle for a sufficiently short interval, giving local uniqueness in the ordinary initial-value sense. A convenient Lipschitz check is a continuous bounded partial derivative f_y on a rectangle: the one-variable mean value theorem gives L=max|f_y|. Failure of this sufficient derivative test alone does not prove nonuniqueness.

Continuity alone need not select one history

Consider the dimensionless equation y′=3|y|^(2/3), y(0)=0. Its right side is continuous for all real y, but it is not locally Lipschitz at y=0: |f(ε)−f(0)|/|ε|=3ε^(−1/3) becomes unbounded as ε decreases to zero through positive values. The local theorem above therefore does not supply uniqueness at this initial point.

Nonuniqueness can be established directly, rather than merely inferred from a failed hypothesis. The constant function y=0 is one solution. For any a≥0, define y_a(t)=0 for t≤a and y_a(t)=(t−a)³ for t≥a. At the joining point both pieces have value zero and derivative zero. Elsewhere direct substitution gives y_a′=3|y_a|^(2/3). Thus every such differentiable function solves the same initial-value problem.

These histories wait at zero for different lengths of time before becoming positive. A slope field shows zero slope at y=0, yet it does not determine the departure time. Initial data plus a continuous rate law do not guarantee uniqueness in every equation. The counterexample identifies the missing control on how rapidly the rate changes with the state near zero.

A failed sufficient hypothesis can also occur in a problem that still has a unique solution. To decide an actual case, provide a direct uniqueness argument or distinct valid solutions; do not equate failure of one theorem with failure of the conclusion. Similarly, checking one nonzero branch after division by a state variable can lose equilibrium solutions. Lesson 138 keeps those excluded cases explicit when deriving solution formulas.

A locally unique solution can still end in finite time

The field f(t,y)=y² is smooth everywhere, so it is locally Lipschitz in y near every finite point. Nevertheless, y′=y², y(0)=1 has solution y=1/(1−t) only up to its finite-time blow-up at t=1 in the forward direction. Its maximal connected classical interval containing 0 is (−∞,1). Smoothness of the field does not bound the solution’s growth for all time.

There is no finite real value y(1) that continuously extends this particular solution across 1. Starting the same formula on the interval (1,∞) gives a separate solution branch, not continuation through the singular time of the initial-value solution. A graphing program that draws both branches without marking the break can suggest a false connection.

A solution can also encounter a boundary of the equation’s stated domain. For y′=1/(1−t), y(0)=0, integration gives y=−ln(1−t) on (−∞,1), and the right side of the equation itself is undefined at t=1. These examples distinguish a singularity of the field from blow-up of a solution while the field remains smooth at every finite state. Both require explicit interval checks.

Stronger global assumptions can yield stronger conclusions. For example, continuity and a uniform global Lipschitz bound in y on a finite time strip support existence and uniqueness across that strip. The local rectangle theorem alone does not provide that global bound. Physical modelling may stop even earlier than mathematical continuation, such as at a tank-emptying event or outside a calibrated temperature range.

Uniqueness and sensitivity are different questions

For y′=ky with the same constant k, two initial values differing by δ produce an exact difference δe^(k(t−t₀)). If k>0, uncertainty grows exponentially; if k<0, it decays forward in time. Both initial-value problems remain unique. A unique answer can therefore be sensitive to the input, especially over long times.

A general forward-time estimate needs stated assumptions. Suppose y and z solve the same scalar equation on [t₀,T], their graphs stay in a common rectangle where f is continuous and Lipschitz in y with constant L≥0, and |z(t₀)−y(t₀)|≤δ. Subtracting their integral equations gives |z(t)−y(t)|≤δ+L∫_(t₀)^t |z(s)−y(s)| ds. It follows that |z(t)−y(t)|≤δe^(L(t−t₀)).

One elementary justification defines W(t)=δ+L∫_(t₀)^t |z−y| ds. Then |z−y|≤W, W(t₀)=δ and W′≤LW. The integrating factor e^(−L(t−t₀)) makes its derivative nonpositive, giving W≤δe^(L(t−t₀)). This proves the displayed bound on the assumed common interval. It does not assert that both solutions stay inside the rectangle or exist until any arbitrarily chosen T.

Continuous dependence means that sufficiently small input changes give correspondingly small output changes on an interval with suitable uniform controls. It is compatible with a large finite amplification factor. Model uncertainty, uncertain parameters and noisy measurements are additional sources of error; a theorem about initial data for a fixed f does not automatically bound them all.

A small residual needs a stability bound and an initial check

For a continuously differentiable approximate curve u(t), define r(t)=u′(t)−f(t,u(t)). Assume |r|≤ε, the initial mismatch is at most δ, and u and a true solution y remain in a common rectangle with Lipschitz constant L on the forward interval. Their integral equations give |u−y|≤δ+∫[L|u−y|+ε] ds. The same integrating-factor argument yields |u(t)−y(t)|≤δe^(LΔ)+ε(e^(LΔ)−1)/L for L>0, where Δ=t−t₀. For L=0 the bound is δ+εΔ.

This estimate requires a residual bound over the entire interval, not a few sampled zeros. It also requires the initial mismatch and Lipschitz/domain conditions. For y′=y, every u=Ce^t has zero residual, but only C=1 satisfies y(0)=1. A zero differential residual alone therefore cannot certify the requested initial-value solution.

A first numerical step comes from the tangent approximation: tₙ₊₁=tₙ+h and wₙ₊₁=wₙ+h f(tₙ,wₙ). This is explicit Euler’s method. It evaluates the rate at the previous approximate state. It is an approximation, not an exact identity for a general solution. If the exact solution is C², its one-step Taylor remainder from an exact starting value is bounded by (h²/2) times a bound for |y″| over that step.

The unscaled one-step defect just described is O(h²) under a uniform second-derivative bound; conventions dividing it by h call the resulting local truncation error O(h). Global error needs accumulation and stability analysis. For y′=−2y, Euler gives wₙ=(1−2h)ⁿw₀: h=1 leaves an alternating sequence of unchanged magnitude although the exact solution decays. A smaller displayed step or plausible-looking plot does not by itself certify accuracy.

Audit the equation, data, theorem and interpretation separately

A complete result records the unknown and its units, independent variable, equation, parameter assumptions, initial data, field domain and connected solution interval. Differentiate a candidate and check the data explicitly. If a formula was obtained by division, square roots or logarithms, track excluded states and branches. A missing equilibrium or a disconnected interval can survive otherwise flawless algebra.

Before applying an existence or uniqueness theorem, name the actual rectangle or domain and verify its hypotheses. When a hypothesis fails, state only that the theorem is inconclusive until a separate argument settles the issue. When a theorem is local, keep its conclusion local. Computational sampling can help locate concerns, but it cannot establish a uniform derivative or residual bound over an uninspected continuum.

The explorer should distinguish exact solutions, derivative residuals, slope-field samples, theorem-admissibility statements and numerical iterates. Its time and state scales need labels. A model that reaches a pole must stop the plotted initial-value branch there; it must not join across the pole or conceal a failed hypothesis behind matching numbers on a few sample points.

Finally, compare the mathematical claim with the physical assumptions used to build it. A unique, numerically well-approximated solution can still describe the wrong physical process. Conversely, the absence of an elementary formula does not make a carefully posed model useless: qualitative arguments, bounds and controlled numerical methods can answer practical questions while preserving their limitations.

Worked examples

Try it yourself

1. What is unknown in y′=t+y: a single number, or a function? What extra datum makes it an initial-value problem?

   Show the answer and reasoning

   The unknown is a function y(t); an initial value y(t₀)=y₀ supplies the datum.

   The equation constrains the function and derivative along an interval. One point is extra information, whose sufficiency for uniqueness still requires appropriate conditions.

2. What are the orders of (y′)²+y=0 and y‴+y=0?

   Show the answer and reasoning

   First order and third order, respectively.

   Order counts derivative order, not an algebraic power of a derivative. Squaring y′ does not produce y″.

3. Is y′+t²y=sin t linear? Is y′=y(1−y) linear?

   Show the answer and reasoning

   The first is linear and nonautonomous; the second is nonlinear and autonomous.

   A coefficient may depend on t in a linear equation, but a y² term is nonlinear in the unknown. Explicit dependence on t is a separate classification.

4. Does u(t)=5e^t solve y′=y, y(0)=2?

   Show the answer and reasoning

   It solves the differential equation but not the initial-value problem.

   u′=u, yet u(0)=5 rather than 2. Both equation and data must be checked.

5. For a regular second-order position equation, what two initial state quantities are normally supplied at one time?

   Show the answer and reasoning

   Position and velocity, q(t₀) and q′(t₀).

   These become the two components of the first-order state system. Local uniqueness also needs the relevant regularity conditions.

6. Why do q″+q=0 with q(0)=q(π)=0 fail to select a unique q?

   Show the answer and reasoning

   Every q=A sin t satisfies those two endpoint conditions.

   The conditions occur at different times and form a boundary-value problem. They do not fix the initial velocity A.

7. For a fixed-volume well-mixed tank, why is outgoing solute mass rate q m/V rather than m/V?

   Show the answer and reasoning

   m/V is concentration; multiplying by volume flow q gives mass per time.

   The dimensions become (solute mass/volume)·(volume/time), matching m′. Model assumptions determine which concentration is carried out.

8. If a constant-ambient temperature difference halves every 8 minutes, what is k in T′=−k(T−T_a)?

   Show the answer and reasoning

   k=(ln2)/8 min⁻¹.

   The difference is multiplied by e^(−8k)=1/2 after 8 minutes. The ambient temperature itself is not the decaying quantity.

9. For N′=rN(1−N/K), define u=N/K and τ=r(t−t₀). What equation does u satisfy?

   Show the answer and reasoning

   du/dτ=u(1−u).

   The factors K and r cancel by the chain rule. Both u and τ are dimensionless; the original scales must be retained when interpreting a plot.

10. What condition on g defines a constant solution of y′=g(y)?

    Show the answer and reasoning

    g(y_*)=0, giving y(t)=y_*.

    The derivative of a constant is zero, so substitution requires the right side to vanish at that state.

11. Write the integral form of y′=f(t,y), y(2)=7 under the continuity assumptions used here.

    Show the answer and reasoning

    y(t)=7+∫_2^t f(s,y(s)) ds.

    The fundamental theorem gives this identity and recovers both the derivative and initial value from it under the stated continuity conditions.

12. On −1≤y≤2, find a valid Lipschitz constant in y for f(t,y)=y².

    Show the answer and reasoning

    L=4 is valid.

    For u,v in that interval, |u²−v²|=|u+v||u−v|≤4|u−v|. Equivalently max|2y|=4.

13. In a rectangle with a=b=1, M=4 and L=4, does h=1/2 meet the contraction proof’s conditions?

    Show the answer and reasoning

    No.

    Mh=2 exceeds b=1 and Lh=2 is not below 1. A smaller h such as 1/8 meets both conditions.

14. If a useful sufficient Lipschitz test fails, may one conclude the initial-value problem has no unique solution?

    Show the answer and reasoning

    No; that theorem is inconclusive.

    Nonuniqueness requires an actual argument such as two distinct valid solutions. Failure of a sufficient condition is not failure of its conclusion.

15. Why does smooth f(t,y)=y² not guarantee y′=y², y(0)=1 exists as a finite real solution for all t?

    Show the answer and reasoning

    Its solution 1/(1−t) blows up at t=1.

    Local smoothness supplies local uniqueness near finite states. It does not impose a global growth bound preventing finite-time blow-up.

16. May one divide t y′=2y by t and invoke a normal-form theorem through t=0?

    Show the answer and reasoning

    No; division excludes t=0.

    At the original initial point, y(0)=0 allows many solutions y=Ct². The divided field 2y/t is undefined there.

17. For y′=2y, how large is the difference at time t≥0 between solutions with initial difference δ? Does growth of that difference contradict uniqueness?

    Show the answer and reasoning

    The difference is δe^(2t); its growth does not contradict uniqueness.

    Uniqueness compares identical data for one equation. Sensitivity compares different data, and may amplify even small differences.

18. For y′=−2y, y(0)=1, explicit Euler with h=1 gives w₁=−1. What does this say about the exact solution?

    Show the answer and reasoning

    It exposes a poor numerical step; the exact solution remains positive, e^(−2t).

    A numerical recurrence has its own stability properties. Its output is not an exact theorem about the underlying initial-value solution.

What to carry forward

• An ODE constrains an unknown function and its derivatives across an interval.
• An initial-value solution must satisfy the equation, initial data and field-domain requirements.
• State a connected solution interval containing the initial time; do not join branches across a pole.
• Order counts derivatives; linearity and autonomy are different classifications.
• Initial state data at one time differ from boundary conditions at separate times.
• Balance laws require explicit modelling assumptions and units consistent with the retained quantity’s derivative.
• A slope field supplies local directions; an equilibrium is checked by substitution.
• Under appropriate continuity, the integral equation is equivalent to the classical initial-value problem.
• Continuity and a uniform local Lipschitz condition give a sufficient local existence and uniqueness theorem.
• Picard iteration can be justified by a contraction estimate, rather than by resemblance of its first few terms to a familiar series.
• Failure of a sufficient hypothesis is inconclusive; a nonuniqueness claim needs distinct valid solutions or another argument.
• A locally unique solution can blow up in finite time even when its field is smooth everywhere.
• Unique solutions can strongly amplify initial uncertainty; sensitivity and nonuniqueness are different.
• Residual bounds control error only with initial-data, domain and stability assumptions over the whole interval.
• Euler iterates are approximations whose numerical dynamics can differ sharply from the exact solution.
• A correct mathematical solution does not by itself validate the physical assumptions of its model.