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For arbitrary real $n\ne 0,1$, work first on an interval where $y>0$. Define exponentiation by $y^r=\exp (r\log y)$, using the real logarithm. The change of variable is differentiable and invertible there, and the chain rule givesMultiplying the Bernoulli differential equation by $(1-n)y^{-n}$ yields the first-order linear differential equationThis is an exact transformation of nonzero solutions of a differential equation on the chosen branch; it is valid far from any reference state and does not discard higher powers as a local linearization would.

Given the initial condition $y(x_0)=y_0>0$, the integrating factor and transformed solution of a differential equation areThe product rule verifies $(\mu z{)}^{\prime }=(1-n)\mu Q$. Recover $y=z^{1/(1-n)}$ only on the connected component of $\{x:z(x)>0\}$ containing $x_0$, inside the original coefficient interval. The first-order linear differential equation can have a longer existence interval than this inverse transformation allows.^[ref]

Negative values need a separately justified real branch. For integer $n$, the required powers are real and differentiable for $y<0$ too, but the inverse must retain the sign selected by the initial condition. For $n=3$, $z=y^{-2}>0$ represents either $y=+1/\sqrt{z}$ or $y=-1/\sqrt{z}$; both square roots are taken of positive $z$. For $n=2$, $z=1/y$ retains the sign directly. For a general noninteger exponent, use the positive branch unless a specific differentiable real negative branch has been established. Never evaluate a real fractional power of a negative value merely because the linear formula has become negative.

Division by $y^n$ excludes $y=0$. When $n>0$ and the original real power is defined at zero, $y\equiv 0$ is an additional solution of a differential equation and must be checked in the original equation. When $n<0$, the equation as written is undefined at zero. On a positive branch the right-hand side is locally Lipschitz in $y$, so the Picard-Lindelöf theorem gives local uniqueness. At zero, the same hypothesis needs a fresh check: it holds for the polynomial $n=2$ example, whereas for $0<n<1$ a nonzero power coefficient prevents local Lipschitz continuity there. If $z\downarrow 0$, the inverse sends $y\to \infty$ for $n>1$, but sends $y\downarrow 0$ for $0<n<1$. The latter endpoint can lead to finite-time extinction in a sublinear Bernoulli equation, where the original equation determines how to join to zero.

A first-order ordinary differential equation with complex independent and dependent variables is solved by a holomorphic function on an ordinary domain, and its analytic continuation can have branch points. A holomorphic right-hand side gives local existence and uniqueness by the integral contraction mapping used in the Picard-Lindelöf theorem.

A direction field for $y^{\prime }=f(t,y)$ places a short segment with slope $f(t,y)$ at each sampled point where $f$ is defined. A function graph of a solution of a differential equation follows these segments: its tangent line has exactly the prescribed slope. Segment length is a drawing choice, rather than a speed; arrows, if used, point toward increasing $t$. An isocline collects points with one common slope.

For Newton's law of cooling, $y^{\prime }=-0.1(y-20)$ with time in minutes and temperature in degrees Celsius, the slopes at $y=80,40,20,10$ are respectively $-6,-2,0,1$ degrees Celsius per minute. Each horizontal row is an isocline, because this autonomous differential equation has no explicit time dependence. Above $20$ the solutions of a differential equation descend, below $20$ they rise, and $y=20$ is an equilibrium point.

The highlighted initial condition $y(0)=80$ selects one curve from the family. The picture reveals approach to the equilibrium point before any algebra is done. Under the hypotheses of the Picard-Lindelöf theorem, distinct function graphs of solutions cannot cross or even meet in the time-state plane. A direction field alone does not guarantee uniqueness: several curves can share a tangent line at the same point when those hypotheses fail. See the delayed-start example under initial value problem.

This geometric interpretation is developed in Direction Fields, Isoclines, Integral Curves.

For a linear group action of $G$ on a state space, an equivariant dynamical system has a vector field satisfyingThus applying a symmetry before computing a velocity gives the same result as applying it to the computed velocity. An equivariant vector field need not be invariant as a vector: its value transforms along with the point where it is evaluated.

If the ordinary differential equation has unique solutions, its flow map obeys ${\Phi }_t(gx)=g{\Phi }_t(x)$ whenever both sides exist. Indeed, $g{\Phi }_t(x)$ has derivative $gF({\Phi }_t(x),\mu )=F(g{\Phi }_t(x),\mu )$ and starts at $gx$, so Picard-Lindelöf theorem identifies it with ${\Phi }_t(gx)$.

Consequently, every subgroup's fixed-point subspace of a group action is invariant under the flow map. If $hx=x$ for all $h$ in the subgroup, then $h{\Phi }_t(x)={\Phi }_t(hx)={\Phi }_t(x)$. This reduces some symmetry-preserving motions to a smaller state space. It also constrains normal forms: the wavevector selection rule for equivariant monomials and the cubic equivariants of the full cube symmetry group are concrete coefficient tests.

Under the Picard-Lindelöf theorem's hypotheses on $U\subset \mathbb{R}\times {\mathbb{R}}^d$, extend a solution of a differential equation as far as possible, to a maximal interval $(T_{-},T_{+})$. If $T_{+}<\infty$, the time-state trajectory $(t,y(t))$ must eventually leave every compact subset of $U$. Equivalently, confinement near $T_{+}$ to one compact subset of $U$ permits continuation past $T_{+}$. See Teschl, Section 2.6, Lemma 2.14 and Corollaries 2.15–2.16, pp. 52–53.

The domain restriction matters even for a bounded trajectory. Restrict $y^{\prime }=1$, $y(0)=0$, to $U=\mathbb{R}\times (-1,1)$. Then $y(t)=t$ exists in this domain only for $-1<t<1$; at $t=1/2,3/4,9/10$ its values are $1/2,3/4,9/10$. As $t\uparrow 1$, it approaches the excluded boundary point $(1,1)$ while remaining bounded. Its trajectory cannot fit into any compact subset of $U$ near that endpoint. This is escape from a finite domain, without finite-time blowup in norm. Enlarging the domain to $\mathbb{R}\times \mathbb{R}$ gives the same formula $y(t)=t$ for every real $t$.

Conversely, $y^{\prime }=y$, $y(0)=1$, stays in the compact set $[0,T]\times [1,e^T]$ for every fixed $T>0$. Its growing exponential function is unbounded over the entire future, but has no finite-time blowup. At $T=2$, its largest value on $[0,2]$ is only $e^2\approx 7.38905610$. The relevant question is boundedness before each finite endpoint, rather than one bound valid forever. The tangent function in the diagram under the Picard-Lindelöf theorem illustrates the other possibility: its norm escapes every bound before $T=\pi /2$.

This criterion is useful because an estimate can secure continuation without an explicit formula. In a finite-dimensional vector space, a bounded closed ball is compact; a bounded state trajectory therefore supplies the needed compact set when its time-state closure is inside the domain. That final proviso prevents mistaking a domain boundary for finite-time blowup.

A common sufficient estimate comes from global Lipschitz continuity in the state. Assume $U=I\times {\mathbb{R}}^d$ for an open time interval $I$, with $f$ continuous and a Lipschitz constant $L\ge 0$ valid for all states on each compact time interval under consideration. For $[t_0,T]\subset I$, let $B={\sup }_{t_0\le s\le T}\Vert f(s,0)\Vert$, finite by continuity. Then $\Vert f(t,y)\Vert \le B+L\Vert y\Vert$, and the Gronwall inequality gives, for $t_0\le t\le T$ wherever the solution already exists,For instance, $y^{\prime }=y+1$, $y(0)=0$, has $L=B=1$ and the exact solution $y(t)=e^t-1$, attaining this estimate; at $t=1$ the value is $e-1\approx 1.71828183$. Finite bounds of this kind give continuation throughout $I$, with the analogous estimate backwards in time. When $I=\mathbb{R}$, this gives existence for all real times. If $I$ has a finite endpoint, it gives no continuation beyond that prescribed domain. The underlying linear-growth criterion is ^[ref].

A brief explanation of the continuation mechanism suffices: within the protecting compact set, the continuous function $f$ is bounded, say by $M$. The integral equation yields $\Vert y(t)-y(s)\Vert \le M|t-s|$ near the finite endpoint, so $y(t)$ has a limit $y_{\ast }$. The protecting compact set puts $(T_{+},y_{\ast })$ inside $U$. Restarting the initial value problem there and using local uniqueness joins the trajectories.

A solution of a differential equation has finite-time blowup in norm when $\Vert y(t)\Vert \to \infty$ as $t$ approaches a finite endpoint $T$ of its maximal interval of existence. This is stronger than merely approaching an excluded boundary of an open state domain while staying bounded. The finite-dimensional continuation criterion distinguishes these obstructions; for a locally Lipschitz continuous ordinary differential equation on $I\times {\mathbb{R}}^d$, a finite maximal endpoint inside $I$ necessarily entails finite-time blowup.

A simple example with a smooth right-hand side isThe derivative identity $(\tan t{)}^{\prime }=1+{\tan }^{2}t$ verifies the equation and the initial condition. Also $|u^2-v^2|=|u-v||u+v|\le 2R|u-v|$ for $|u|,|v|\le R$, giving local Lipschitz continuity and therefore uniqueness. Yet the positive blowup time is $T=\pi /2\approx 1.57079633$. Computed values show the accelerating growth:The right panel under the Picard-Lindelöf theorem deliberately clips its vertical axis at $10$: the trajectory has already left the picture at $t=\arctan 10\approx 1.47112767$, before reaching $T$. A plot ceiling is a display limit, not a finite value at blowup.

The tangent function also quantifies the last part of the trajectory. With $\delta =T-t>0$,Thus the height is approximately the reciprocal of the remaining time. At remaining times $\delta =0.1,0.01,0.001$, the heights are approximately $9.96664442,99.99666664,999.99966667$. These follow from the sine and cosine limits at zero and explain the steep rise without relying on the appearance of a clipped function graph.

The distinction is useful when interpreting a solution of a differential equation: uniqueness certifies a trajectory while it exists, while growth estimates decide how far it can be followed. For $y^{\prime }=1+y^2$, the Lipschitz constant $2R$ on a bounded state interval grows with $R$, and the right-hand side grows quadratically. There is no global Lipschitz constant or linear-growth bound of the kind used by the finite-dimensional continuation criterion. Smoothness at every finite state is entirely compatible with finite-time blowup.

A fundamental system for a homogeneous linear ordinary differential equation of order $n$ on an interval is a basis of its vector space of homogeneous solutions. Every solution has a unique expression ${\sum }_{j=1}^n{c}_j{y}_j$ with constant coefficients. Under the usual continuity assumptions and a nonvanishing leading coefficient, a nonzero Wronskian at one point proves that $n$ solutions form such a system: their initial-value vectors are independent and span all possible initial data, and the Picard-Lindelöf theorem supplies uniqueness. For a first-order vector system, the corresponding columns form a fundamental matrix.

If $H$ is $C^2$ and its spatial Hessian matrix is uniformly bounded on every finite time interval, its Hamiltonian vector field is globally Lipschitz continuous in phase space with at most linear growth. The Picard-Lindelöf theorem and Gronwall inequality give a unique Hamiltonian flow for all finite forward and backward times. Bounds on ${\nabla }_{z}H(t,0)$ follow from time continuity on compact intervals.

The displayed implicit ordinary differential equation has a singular initial point at the origin. On a nonconstant regular branch, treating $x$ as a function of $y$ gives $dx/dy+x\cot y=2\cos y$. The integrating factor $\sin y$ yields the first integral of an ordinary differential equationOne can verify this identity without dividing by $\sin y$: its derivative along a solution is $\sin y-(2\sin y-x)\cos y{y}^{\prime }=0$.

The initial point forces $C=0$, so $\sin y(x-\sin y)=0$. Near the origin this gives the two differentiable solutions $y=0$ and $y=\arcsin x$, with initial slopes zero and one. A division by $\sin y$ would lose the first solution. A piecewise switch between the two at zero would fail differentiability because their one-sided slopes differ.

There is no conflict with the Picard-Lindelöf theorem: solving explicitly for $y^{\prime }$ requires division by $2\sin y-x$, which vanishes at the initial point. A regular implicit equation or a local uniqueness theorem must be checked before claiming a unique solution through a singular point.

An initial value problem combines an ordinary differential equation with an initial condition, for example $y^{\prime }=f(t,y)$ and $y(t_0)=y_0$. The task is to find a solution of a differential equation on an interval containing $t_0$ which has that starting value. For a continuous linear differential equation $u^{\prime }=A(t)u$, every prescribed finite vector $u_0$ determines a unique solution on the coefficient interval. A boundary value problem can instead prescribe values at distinct endpoints.

For the cooling initial value problem $y^{\prime }=-k(y-T_a)$, $y(t_0)=y_0$, where $k>0$ and $T_a$ are constant, the unique solution isTake $T_a=20$ degrees Celsius, $y_0=80$ degrees Celsius, $t_0=0$, and $k=0.1$ per minute. Then $y(10)=20+60e^{-1}\approx 42.07$ degrees Celsius. Reaching $30$ degrees Celsius requires $60e^{-0.1t}=10$, so $t=10\log 6\approx 17.92$ minutes. The time constant $1/k=10$ minutes reduces the excess temperature by a factor $e^{-1}$; it does not reduce the absolute temperature by that factor.

Uniqueness needs assumptions. If $f$ is a continuous function on an open neighborhood of $(t_0,y_0)$ and satisfies a local Lipschitz condition in $y$, uniformly in $t$, the Picard-Lindelöf theorem gives a unique local solution. Concretely, a finite $L$ bounds $|f(t,u)-f(t,v)|\le L|u-v|$ in a sufficiently small rectangle. Cooling satisfies this with $L=k$. Local existence does not promise existence for all future time: $y^{\prime }=y^2$, $y(0)=1$, has $y(t)=1/(1-t)$ on $(-\infty ,1)$ and finite-time blowup at $t=1$.

Continuity alone is insufficient for uniqueness. Consider the dimensionless initial value problem involving a square root and an absolute value:Besides $y(t)\equiv 0$, each delay $a\ge 0$ gives a solution for $t\ge 0$:At $t=4$, delays $a=0,1,2$ give $y=4,9/4,1$, despite identical starting data. These are valid solutions of a differential equation: for $a>0$, both one-sided derivatives at the join are zero; for $a=0$, the right derivative at the starting time is zero. After the join, $y_a^{\prime }=(t-a)/2=\sqrt{|y_a|}$ because $t>a$. Squaring alone would also suggest a parabola before the join, but its negative derivative there would violate the equation. Since $y^{\prime }\ge 0$, a forward solution starting from zero stays nonnegative.

The right-hand side is a continuous function but fails the Lipschitz condition at zero: for $h>0$, $|\sqrt{|h|}-0|/|h|=1/\sqrt{h}$ is unbounded. Thus the Peano existence theorem still supplies local existence, while the Picard-Lindelöf theorem does not supply uniqueness. Related waiting-time examples and a sufficient uniqueness condition appear in MIT 18.034, Lecture 2: Fundamental Principles.

A brief proof outline explains the mechanism: rewrite the initial value problem as the integral equation $y(t)=y_0+{\int }_{t_0}^{t}f(s,y(s))ds$. On a short enough interval, the Lipschitz condition makes the right-hand side a contraction mapping on a closed set of continuous functions near $y_0$. The contraction mapping theorem gives a unique fixed point; the fundamental theorem of calculus turns it back into a solution.

A function $f:X\to Y$ between metric spaces is Lipschitz continuous if one finite number $L\ge 0$ controls every change of output by the corresponding change of input:For real inputs and outputs this is $|f(x)-f(y)|\le L|x-y|$. The same bound must work for every pair in the specified domain of a function. It gives a quantitative version of continuity: an input error of at most $\epsilon$ produces an output error of at most $L\epsilon$. The least possible bound is the Lipschitz constant. Lebl presents the real and metric-space definitions and their connection with uniform continuity. ^[ref]

Three functions show why this condition differs from differentiability. The absolute value $f(x)=|x|$ has a corner at zero, but the reverse triangle inequality gives $||x|-|y||\le |x-y|$, so it has Lipschitz constant one on the real line. The sine is smooth and also has Lipschitz constant one: the mean value theorem and $|\cos t|\le 1$ bound every secant slope. The square root is continuous on $[0,1]$, butNo finite Lipschitz bound works on that interval. At $h=1,1/4,1/16,1/64$, the ratios are $1,2,4,8$. Smaller input changes near zero therefore require larger amplification factors.

For a real function, Lipschitz continuity means that the absolute slopes of all secant lines are bounded. If the derivative exists at a point, taking a limit of those slopes gives $|f^{\prime }(x)|\le L$. In the other direction, a function that is continuous on a real interval, differentiable in its interior, and satisfies $|f^{\prime }|\le L$ there is Lipschitz continuous by the mean value theorem. The interval hypothesis matters: the theorem uses the segment connecting the two inputs.

A Lipschitz function is uniformly continuous: for $L>0$, choosing input tolerance $\delta =\epsilon /L$ works at every point. If $L=0$, the function is constant. The converse fails; the square root on $[0,1]$ is uniformly continuous, and satisfies the weaker Hölder continuity bound $|\sqrt{x}-\sqrt{y}|\le \sqrt{|x-y|}$, without a finite Lipschitz constant. These distinctions are developed in Lebl's worked examples. ^[ref]

The usefulness is controlled sensitivity. Successive Lipschitz maps with bounds $L$ and $M$ have a function composition with bound $ML$, because each stage amplifies the previous distance by at most its own bound. For an ordinary differential equation, Lipschitz continuity in the unknown state is the condition that makes nearby trial solutions controllably close in the Picard-Lindelöf theorem. For incomplete data, the McShane extension theorem fills in unobserved values while preserving the same bound. This one inequality connects error estimates, uniqueness, and interpolation.

A function between metric spaces is locally Lipschitz if every point has a neighbourhood on which one finite Lipschitz bound works for every pair. The neighbourhood and bound may depend on the point. A continuously differentiable real function on an open interval has this property: continuity of its derivative bounds that derivative on a sufficiently small closed interval, and the mean value theorem bounds every secant slope there.

The function $x^2$ is locally Lipschitz at every real point but has no single bound on the real line. The square root is locally Lipschitz on $(0,\infty )$, and has the bound $1/(2\sqrt{a})$ on $[a,b]$ for $0<a<b$. Adding zero to its domain changes the conclusion at that point: the ratios in the Lipschitz continuity example become unbounded, so it is not locally Lipschitz at zero. This distinction is what permits the delayed, nonunique ODE solutions illustrated under Picard-Lindelöf theorem.

For the ordinary differential equation $y^{\prime }=3x^2/(4y^3)$ through $y(0)=b>0$, consider a symmetric existence rectangle $|x|\le h$, $|y-b|\le k<b$. Its maximum slope bound is $M=3h^2/[4(b-k{)}^3]$. The condition $Mh\le k$ becomes $h^3\le 4k(b-k{)}^3/3$. Differentiating $k(b-k{)}^3$ shows its maximum occurs at $k=b/4$, givingThis is the largest interval guaranteed by this particular centered-rectangle criterion. Direct separation of variables instead gives $y(x)=(b^4+x^3{)}^{1/4}$ on the maximal real interval $x>-b^{4/3}$. The comparison shows the difference between a sufficient local existence theorem and the actual continuation interval. The endpoint obstruction is the zero of the denominator at $y=0$.

For a first-order ordinary differential equation rational in the dependent variable, with coefficients meromorphic on the chosen independent-variable surface, analytic continuation along an arc ending outside the fixed exceptional set has a definite limit in the Riemann sphere. Its movable singularities are poles or algebraic branch points. To see the limiting assertion, the cluster set of a continuous lifted arc is connected and compact. Any cluster value where the vector field is holomorphic permits uniform local continuation by the Picard-Lindelöf theorem. Otherwise the cluster set lies in finitely many coefficient poles, so it is a singleton. At a pole of the vector field, the inverse equation is holomorphic; its solution has a finite-order critical point, giving a convergent root expansion. The reciprocal dependent variable handles infinity. Points where both reduced numerator and denominator vanish belong to the fixed exceptional set and are excluded from this argument.

The Picard-Lindelöf theorem gives local existence and uniqueness for an initial value problem when $f$ is continuous on a rectangle around $(a,b)$ and is Lipschitz continuous in its second variable there, with a constant uniform in the first variable. A sufficient hypothesis is that $f$ and $\partial f/\partial y$ are continuous on that rectangle.

More quantitatively, if the rectangle is $|x-a|\le h_0$, $|y-b|\le k$, and $|f|\le M$, a sufficiently small $h>0$ with $h\le h_0$ and $hM<k$ yields a unique differentiable solution on $[a-h,a+h]$ remaining inside the vertical strip. The proof uses the integral equationand Picard iteration, or the Banach fixed-point theorem after shrinking $h$ to make the integral operator a contraction. The continuity and Lipschitz hypotheses must hold near the actual initial point.

Take any $a\ne 0$ and any $b\in \mathbb{R}$. Choose $0<h_0<|a|$, so $[a-h_0,a+h_0]$ avoids zero. Onboth $f$ and $f_y=(1-2x)/x^2$ are continuous. They are bounded by compactness, and the bound on $f_y$ gives the uniform Lipschitz continuity in $y$ required by the Picard-Lindelöf theorem.

If $M$ is a bound for $|f|$, shrink $h$ so $0<h<h_0$ and $hM<1$. Then the solution has $|y(x)-b|\le hM<1$, which supplies even strict inequalities at both edges of the requested vertical strip.

At $a=0$, neither $f$ nor the original exponential expression is defined. Thus the theorem applies exactly for $a\ne 0$, with no restriction on $b$. Failure of the hypotheses alone would not prove nonexistence; the explicit solutions below settle that issue.

For fixed $a\ne 0$ and $B>0$, the rectangleis compact and stays away from $x=0$. The function $f$ in part (i) is continuous there. The extreme value theorem makes $|f|$ attain a finite maximum $M_0$ on $K$; choosing $\boxed{M=M_0+1}$ gives the requested strict bound.

The absolute values make this a positive-$x$ rectangle even when $a<0$. Its boundedness is still true, but for a negative initial point the Picard-Lindelöf theorem must instead be applied to a rectangle centered at that negative point, as in part (iii).

For every $a\ne 0$ and every real $b$, there is exactly one solution on the maximal connected interval containing $a$. The formula in part (i) proves existence on the entire appropriate half-line, and the initial condition fixes its sole constant uniquely. This agrees with the local Picard-Lindelöf theorem.

For $a=0$, there is no classical solution on an interval containing the initial point, whatever $b$ is: the printed equation is not defined there. Even if one only imposes the equation for $x\ne 0$ and asks for a finite continuous extension across zero, the negative side rules it out. For $x\to 0^{-}$:

Thus no choice of the negative-side integration constant admits a finite left limit.

There is a useful distinction with a separately extended one-sided problem. On $x>0$, every member of the solution family tends to zero with all its derivatives as $x\downarrow 0$, since the exponential dominates every inverse power of $x$. If the forcing is separately defined to be zero at the right endpoint, that modified one-sided problem has infinitely many smooth solutions with $y(0)=0$, and none with $y(0)\ne 0$. Those one-sided extensions do not give solutions on the two-sided intervals in the question.

Work with the L1 norm on ${\mathbb{R}}^2$. The scalar Lipschitz hypothesis impliesWe may take $L\ge 0$. The compact set $S$ and continuity give a finite bound $M={\max }_S\Vert \mathbf{f}{\Vert }_1$. ChooseLet $I=[a-\eta ,a+\eta ]$. The space $C_{\eta }$ of continuous maps $I\to B_k(\mathbf{b})$ is complete in the supremum norm: it is a closed subset of the Banach space $C(I,{\mathbb{R}}^2)$. Define the Picard operator on a closed ball byFor any $\mathbf{g}\in C_{\eta }$, the integral triangle inequality givesThe integrand is continuous, so the result is continuous and remains in the prescribed closed ball. Thus $\mathcal{P}$ is a self-map of the complete space. Moreover,It is a contraction mapping. The contraction mapping theorem gives a unique fixed point $\mathbf{y}$. By the fundamental theorem of calculus, the fixed-point identity makes $\mathbf{y}$ continuously differentiable and gives ${\mathbf{y}}^{\prime }=\mathbf{f}(x,\mathbf{y})$ and $\mathbf{y}(a)=\mathbf{b}$. Its first component is twice continuously differentiable, because its derivative is the second component.

Conversely, integrating any solution that remains in the ball makes it a fixed point, so it equals $\mathbf{y}$. A solution starting at $\mathbf{b}$ cannot first reach the boundary of this ball during $I$: before such an exit the same estimate gives distance at most $\eta M<k$, contradicting distance $k$ at the first exit. This proves existence and uniqueness on $|x-a|\le \eta$, not merely formal convergence of an iteration. It is the local Picard-Lindelöf theorem construction.

Near $x=0$, separation of variables gives $dy/y=dx/(1-x)$, so $\log |y|=-\log (1-x)+C$. The initial condition selectsDirect substitution verifies both the ordinary differential equation and the initial condition. The endpoint $x=1$ is excluded because the coefficient and this solution are singular there; the displayed interval is the maximal connected real interval containing the initial point.

For $|x|\le h<1$ and any two admissible values $y,\tilde{y}$,This is a Lipschitz condition in the dependent variable, uniform in $x$, with Lipschitz constant $L=(1-h{)}^{-1}$. Also $|f(x,y)|\le (1+k)/(1-h)$ on the prescribed rectangle, so the usual local hypotheses of the Picard-Lindelöf theorem hold.

Integrating from the initial point gives the integral equationConversely, differentiating this equation using the fundamental theorem of calculus recovers the initial value problem. Start Picard iteration with the constant initial approximation $y_0(x)=1$ and setThe first three approximations areTo prove the general expression by mathematical induction, put $\ell (x)=\log (1-x)$. The initial approximation gives the formula for $n=0$. If it holds for $n$, then the substitution rulegivesThus the formula follows at every stage.

For every fixed real $x<1$, these are partial sums of the exponential series evaluated at the finite real number $-\log (1-x)$. ThereforeThis range is larger than $|x|<1$: the expansion is in powers of $\log (1-x)$, not a Taylor series in $x$. For example it converges at $x=-2$. It also converges uniformly on every compact interval contained in $(-\infty ,1)$, because $|\log (1-x)|$ is bounded there and the exponential series supplies a uniform summable bound. At $x=1$ the approximations are undefined, while for $x>1$ the real integral from zero crosses the singularity. This is Picard iteration for a simple-pole linear differential equation.