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This initial value problem uses a square root and absolute value to define a continuous function on all of ${\mathbb{R}}^2$. For every finite waiting time $c\ge 0$, a forward solution of a differential equation on $t\ge 0$ isThe constant solution of a differential equation $y_{\infty }(t)=0$ is another choice. For example, at $t=3$, waits $c=0,1,2,3$ give values $9,4,1,0$. These are dimensionless mathematical trajectories; no measured physical waiting time is being supplied.

Verification uses the original ordinary differential equation, including the join. Before departure, both $y_c^{\prime }$ and $2\sqrt{|y_c|}$ vanish. For $t>c$, their common value is $2(t-c)$. For $c>0$, the difference quotient at the join isThus the derivative exists at $c$, equals the rate law there, and is continuous across the join. For $c=0$ the same computation gives the right derivative at the starting endpoint. For $c>0$, the left and right second derivatives at departure are $0$ and $2$: these forward solutions of a differential equation are continuously differentiable functions, without being twice differentiable at departure. The local two-sided statement of the Peano existence theorem is separate from this explicitly forward family.

The rate law tends to zero as $y\to 0$, but its secant slopes from zero are unbounded:No finite local Lipschitz constant exists at zero. At $\epsilon =10^{-2},10^{-4},10^{-6}$ the quotient is $20,200,2000$, even though the rates are shrinking to zero. Continuity controls the rate itself; Lipschitz continuity would control its change relative to the state displacement.

There is also a numerical lesson. The forward Euler method starting exactly at zero gives $Y_{k+1}=Y_k+2\Delta t\sqrt{|Y_k|}=0$ at every step. It reproduces the constant solution of a differential equation for every positive step size, while the moving trajectories remain equally valid. A computation producing one trajectory cannot establish uniqueness. Likewise, the separable differential equation calculation that divides by $\sqrt{|y|}$ only describes the moving pieces: checking the original equation recovers the zero pieces and their joins.

A homogeneous first-order differential equation has the form $y^{\prime }=F(y/x)$ on a domain where $x\ne 0$. The substitution $v=y/x$, so $y^{\prime }=v+x{v}^{\prime }$, reduces it toWhere $F(v)-v\ne 0$, this is a separable differential equation, with $\int dv/[F(v)-v]=\ln |x|+C$. Values $v_0$ satisfying $F(v_0)=v_0$ give the additional straight-line solutions $y=v_{0}x$, which must be checked before division. This scaling-based use of “homogeneous” is different from a homogeneous linear differential equation.

For example, $y^{\prime }=y/x+x/y$ gives $F(v)=v+1/v$, so $v{v}^{\prime }=1/x$ and $y^2=x^2(2\ln |x|+C)$. Real explicit branches are $y=\pm x\sqrt{2\ln |x|+C}$ on intervals where the expression under the square root is positive. Zeros of a denominator and endpoints of a branch must still be checked in the original ordinary differential equation.

Newton's law of cooling models an object's temperature $y$ using a positive constant rate $k$ and ambient temperature $T_a$. The displayed ordinary differential equation assumes one temperature represents the object, the environment remains at $T_a$, and the proportional rate stays fixed over the modeled range. It also describes warming when $y<T_a$. These are modeling assumptions: ^[ref] explain the proportional law through a tangent line approximation and caution about large temperature differences.

Dimensional analysis requires $k$ to have units of inverse time. The product $k(t-t_0)$ in the exponential function solution is therefore dimensionless. The equilibrium point $y=T_a$ attracts both warmer and colder starting states: the excess temperature undergoes exponential decay. See initial value problem for computed cooling times, direction field for the family of trajectories, and separable differential equation for a derivation.

Near $(1,1)$ the solution of a differential equation has $y>0$, so the separable differential equation givesEquivalently its trajectory lies on the level setOn $y>0$, the function $G(y)=y-\log (1+y)$ is strictly increasing because $G^{\prime }(y)=y/(1+y)>0$, and its range is $(0,\infty )$. Thus the positive branch exists exactly whereThe connected component containing $x=1$ isThe implicit function theorem gives a smooth solution of a differential equation at every such $x$. It remains finite at every finite $x>x_{\ast }$, because $G$ increases without bound. As $x\downarrow x_{\ast }$, we have $y\downarrow 0$, and the original right side $x+x/y$ becomes singular. Hence this is the maximal interval $(x_{\ast },\infty )$ through the prescribed initial condition.

In particular, the solution is not defined for all $x\in (0,2)$. It cannot pass through $x=x_{\ast }$ as a classical solution, since its required value there is $y=0$.

Where $y\ne 0$ and $x^2\ne 1$, the separable differential equation givesThe requested point selects the implicit solution $2y^2+1=(x^2-1{)}^2$. The two real branches approaching that point areThere is an important initial value problem qualification: the differential equation as printed has no classical solution through the specified initial point. Its right side is undefined at $y=0$, and these branches have an infinite one-sided derivative as $x\downarrow \sqrt{2}$. They satisfy the differential equation on $x>\sqrt{2}$ and extend continuously to the requested point. Multiplying by the vanishing denominator produces an implicit differential equation with a larger domain; it does not repair the original initial value problem.

Set $u=x+y$, so $u^{\prime }=1+y^{\prime }=1+u^2/2$. The new equation is a separable differential equation:Absorbing the constant scaling into $C$, the real solutions areEach solution is taken on a connected interval avoiding its tangent poles. No constant real solution for $u$ was lost by separation, because $1+u^2/2$ is strictly positive for real $u$.

The right-hand side depends on $x+y$, suggesting the substitution $u=x+y$. Then $y^{\prime }=u^{\prime }-1$, so the ordinary differential equation becomesThe original equation is undefined at $u=0$. If $u\ne 2$, this is a separable differential equation:Substituting back gives the implicit familyDifferentiating the implicit equation gives $y^{\prime }-1+2(1+y^{\prime })/(x+y-2)=0$, or $(x+y)y^{\prime }=x+y-4$, verifying it on its permitted intervals.

Division by $u-2$ excluded the constant solution $u=2$, which must be restored separately:It satisfies $y^{\prime }=-1=(2-4)/2$. Together these give all local real solutions on intervals where $x+y\ne 0$.

On an interval with $x\ne 0$, this is a homogeneous first-order differential equation. Put $y=xv(x)$, so $y^{\prime }=v+x{v}^{\prime }$. Substitution and division by $x^2$ giveThe constant choices $v=0$ and $v=2$ give $y=0$ and $y=2x$. Otherwise the separable differential equation givesAbsorbing the sign and the exponential integration constant yields $(v-2)/v=C{x}^2$. Relabelling the arbitrary constant then givesThe first family already includes $y=2x$ at $K=0$. Choose any interval where its denominator is nonzero. These give all solutions on intervals avoiding $x=0$. Both families extend through $x=0$ with $y(0)=0$, as required by the original ordinary differential equation. Because it is singular there, differentiable continuations of the nonzero family may use different $K$ on the two sides: both one-sided derivatives at zero equal $2$.

This is a separable differential equation. On the solution with $y<1$, integration from the initial condition givesThe right side lies in $[0,1)$, so taking its sine stays on the required branch. The solution isNear the origin, $1-\sqrt{1-x^2}=x^2/2+O(x^4)$, so $y=x^2/2+O(x^4)$ and the tangent line is horizontal. At the other end, $y\to \sin 1$, whileThus the curve has a vertical limiting tangent line as $x\uparrow 1$, even though $x=1$ is excluded from the domain.

For the direction field, the slope is nonnegative everywhere in the square. The segments are horizontal on $x=0$ and on $y=1$, become steeper as $x$ increases at fixed $y<1$, and become flatter as $y$ increases at fixed $x$. Near $x=1$ they are almost vertical unless $y$ is correspondingly close to one. The excluded corner $(1,1)$ has no single limiting slope: the ratio $(1-y^2)/(1-x^2)$ depends on the path of approach.

For $r>0$, the polar coordinates satisfyA positive constant radius must therefore be one. The nonconstant periodic solutions arewith least period $2\pi$ and counterclockwise orientation. The origin is also a stationary solution, hence periodic in the degenerate sense of a constant function; it is a point rather than a nontrivial closed orbit. There are no other constant-radius solutions.

For $r_0>0$, put $v=r^2$. Then $\dot{v}=2v(1-v)$, a logistic differential equation. Solving this separable differential equation givesThe denominator is positive for every forward time. When $0<r_0<1$, the radius increases monotonically to one; when $r_0>1$, it decreases monotonically to one; and when $r_0=1$ it stays there. Thus every nonzero orbit approaches the unit circle while continuing to rotate. More precisely, its distance from the phase-matched circular solution tends to zero; its Cartesian coordinates do not converge to a single point. For $r_0=0$, the solution stays at the origin.

The origin is the only equilibrium point. Its Jacobian matrix iswhose eigenvalues are $1\pm i$. It is an unstable focus. The unit circle is an attracting limit cycle, rather than a curve of equilibria: on it the vector field remains nonzero. Linearizing the radial equation at $r=1$ gives $\dot{\delta r}=-2\delta r+O(\delta r^2)$, confirming radial attraction. All other radii are strictly monotone, so there are no additional nontrivial periodic orbits.

Before time one, both forcing terms vanish. The separable differential equation ${\theta }_i^{\prime }=-a{\theta }_i$ gives ${\theta }_i(t)=C_i{e}^{-at}$. The initial values give $C_1=C_2=T_0-T_{\infty }$, and henceIn particular the common left-hand limit at time one is $T_{\infty }+(T_0-T_{\infty })e^{-a}$.

A passive disk is heated predominantly by intercepted stellar radiation, rather than by its own viscous dissipation. A rough luminosity criterion, ignoring geometric factors, isThe intrinsic accretion disk luminosity is of order $GM\dot{M}/(2R_{\ast })$, with additional accretion energy potentially released at the stellar surface. The local test is more precise: absorbed irradiation must dominate the viscous flux at the radius being considered. For an ordinary steady Keplerian accretion disk, sufficiently far outside its inner edge,where $\alpha$ is the irradiation grazing angle of a disk. Reflection, very small incidence angles and shadowing can invalidate a simple total-luminosity comparison. The following calculation assumes an illuminated, absorbing surface.

At a point $(R,z_s)$ on the upper surface, the outward unit normal in the meridional plane is proportional to $(-z_s^{\prime },1)$, and a light ray from the point source is directed along $(R,z_s)$. Its flux through the upper face is the incident inverse-square flux times minus their scalar product:For a thin disk with small surface slope this becomesThe tangent slope and ray slope differ by $\alpha \simeq z_s^{\prime }-z_s/R$. Positive incidence requires disk flaring, $d(z_s/R)/dR>0$, and an unobstructed ray from the star. A negative value of the formula would mean that direct illumination of this face is absent, not that it receives a negative heating flux. The hierarchy $z_s\gg R_{\ast }$ makes the finite angular size of the star negligible here.

Let $a=\mathcal{R}/\mu$, so the ideal gas equation of state gives the isothermal sound speed $c_s^2=a{T}_c$. In the stellar vertical gravitational potential, hydrostatic equilibrium has vertical acceleration $-{\Omega }_K^{2}z$, where ${\Omega }_K^2=GM/R^3$. Integration of $c_s^2{\partial }_z\rho =-\rho {\Omega }_K^{2}z$ gives a Gaussian atmosphere with disk scale heightConsequently $q=z_s/R=fH/R$ satisfies $T_c=GM{q}^2/(a{f}^{2}R)$. Substitute this into the stipulated Stefan–Boltzmann law $F_d=\sigma T_c^4$ and the irradiation flux:Here $f$ specifies surface height in units of disk scale height and is unrelated to the wind-energy fraction in the preceding solution. This separable differential equation integrates exactly toIf the thin-disc solution is formally matched to $q(R_M)=1$, then $B=1-R_0^2/R_M^2$, givingIt follows that the flaring passive disk asymptotic isThis applies in the thin intermediate region $R\ll \min (R_0,R_M)$, in addition to the point-source and stellar-gravity assumptions. The corrections are of relative order $R^2/R_0^2+R^2/R_M^2$. In particular, the printed hierarchy $R\ll R_M$ must still be accompanied by the initial thin-disc condition $q\ll 1$: if $R_M\gg R_0$, radii larger than $R_0$ cannot satisfy that approximation. The matching at $q=1$ is an extrapolation used to set an integration constant; the derivation is not a valid treatment of a thick surface at $R_M$ itself.

Identify the stellar surface temperature with its effective radiating temperature, so that luminosity of a spherical blackbody gives $L_{\ast }=4\pi R_{\ast }^2\sigma T_{\ast }^4$. The given stellar surface scale height has $H_{\ast }^2=a{T}_{\ast }{R}_{\ast }^3/(GM)$, hence $GM/(a{T}_{\ast }{R}_{\ast })=(R_{\ast }/H_{\ast }{)}^2$. Substituting into $R_0$ yieldsThus a small stellar surface scale height leaves a wide interval above $R_{\ast }$ and below $R_0$ in which the thin disk flaring law can apply. The point-source approximation imposes the stronger lower cutoff obtained from $z_s\gg R_{\ast }$:This cutoff is still much smaller than $R_0$ when $R_0\gg R_{\ast }$. The full interval is therefore $R_{\ast }^{7/9}{R}_0^{2/9}\ll R\ll \min (R_0,R_M)$, whenever these bounds overlap. An assumed thickening radius must lie above the lower cutoff for all of the original approximations to hold simultaneously. The stellar luminosity-temperature relation is essential to the final comparison; independent arbitrary values of $L_{\ast }$ and $T_{\ast }$ would not imply it.

The equilibrium points are $y=0,1,-1$. To obtain the other solutions of this separable differential equation without losing their signs, set $z=y^{-2}$ on an interval where $y\ne 0$. Then $z^{\prime }=1-z$, so $z=1+K{e}^{-t}$. Therefore all real solutions areon intervals where $1+K{e}^{-t}>0$. This family includes $y\equiv \pm 1$ when $K=0$. For a prescribed finite nonzero $y(0)=y_0$, $s=\mathrm{sgn}{y}_0$ and $K=y_0^{-2}-1>-1$, so the solution exists for all future time. The cubic logistic differential equation hasThese are the only possible limits. The sign cannot change by uniqueness, and the denominator tends to one. A nonzero initial condition producing a constant solution is $y(0)=1$; $y(0)=-1$ also works. The central equilibrium point is unstable and the two outer equilibrium points attract solutions from their respective half-lines.

For $x\ne \pm 1$, eliminating time gives the separable differential equationand henceDefine the possible limit at $x=s\in \{-1,1\}$ for a trajectory starting off the boundary lines byThe $k<0$ set is empty because $|y|\to \infty$, so there is no finite accumulation point. Since $\dot{x}<0$ on $(-1,1)$ and $\dot{x}>0$ outside $[-1,1]$, the complete classification off the boundary lines isFor $x_0>1$ the solution has forward finite-time blow-up, and for $x_0<-1$ it has backward finite-time blow-up, explaining the other empty entries under the usual infinite-time definitions.

It remains to treat $x_0=s=\pm 1$. Here $x(t)=s$ and $y(t)=y_0{e}^{kst}$. If $y_0=0$, both limit sets equal $\{(s,0)\}$. If $y_0\ne 0$, thenThis includes all boundary cases in both $k$ and the initial point.

For a component with the barotropic equation of state $P=w\rho$, the cosmological perfect-fluid continuity equation isUsing $H=\dot{a}/a$ makes this a separable differential equation:Taking $a_0=1$ gives the constant-equation-of-state density scalingThe spatially flat Friedmann equation and the cosmological density parameters therefore giveAlong a radial light ray, $d\chi =dt/a$. Since the redshift-time relation is $dt=-dz/[(1+z)H]$ and $a=(1+z{)}^{-1}$,Hence the comoving radial distance is

Use the change of variables $u=x+y$. Then $u^{\prime }=1+y^{\prime }=1+\cos u=2{\cos }^2(u/2)$, a separable differential equation. Where $u$ is not an odd multiple of $\pi$, separation of variables givesIndeed the derivative of $\tan (u/2)$ with respect to $u$ is $1/(1+\cos u)$. The nonconstant branches can also be written $y=2\arctan (x+C)+2k\pi -x$, $k\in \mathbb{Z}$. Dividing by $1+\cos u$ excluded its zeros: the missing equilibrium solutions for $u$ give the additional solutions of a differential equationThey satisfy $y^{\prime }=-1=\cos (x+y)$ directly. The transformed right side is continuously differentiable, so the uniqueness theorem for ordinary differential equations prevents a nonconstant branch from crossing an equilibrium solution at finite $x$.

Work with $t>0$ and write $u=t^{-1/2}{e}^{-x^2/(4t)}$. The chain rule givesThus $u_{xx}=u_t$: it is a constant multiple of the one-dimensional Gaussian heat kernel.

For the modified equation, put $f=g(t)u$. Substitution gives $g{u}_{xx}=g^{\prime }u+g{u}_t-gu/t$. The heat-equation terms cancel, and $u>0$ permits division to obtain $\boxed{g^{\prime }=g/t}$. Solving this separable differential equation givesThe zero solution is included by $C=0$. This is the scalar integrating factor for a reaction-diffusion equation with reaction coefficient $1/t$, because the modified equation is $f_t=f_{xx}+f/t$. Its sign produces growth. Indeed ${\int }_{\mathbb{R}}udx=2\sqrt{\pi }$ is constant, while ${\int }_{\mathbb{R}}fdx=2C\sqrt{\pi }t$ increases linearly for $C>0$; neither profile is being assigned unit normalization here.

The separable differential equation gives $d(1-y)/(1-y)=-3dx/x$. Thus $1-y=C{x}^{-3}$ on $x>0$, and the initial value gives $C=1$. HenceThe initial-value branch cannot cross its singularity at zero.

Away from $x=-1$ the separable differential equation givesOn the interval containing $x=1$, one has $x>-1$, so the positive initial value selects $y=K(1+x^3{)}^{1/3}$. Substitution of $y(1)=2$ gives $K=2^{2/3}$. ThusThe zero solution omitted by division by $y$ cannot satisfy the initial value. Although the selected solution tends to zero as $x\downarrow -1$, its derivative diverges there; it does not extend through that point as a classical differentiable solution.

For this separable differential equation, away from $x=1$, separation gives $\ln |y+1|=\ln |x-1|+C$, or $y+1=K(x-1)$ on a connected interval. The initial condition gives $2=-K$, henceThe maximal interval containing $x=0$ on which this solves the printed differential equation is $\boxed{-\infty <x<1}$. Although the linear formula extends through one, the right-hand side there is $0/0$ and the original equation is undefined. The equilibrium solution $y=-1$, lost if one divides by $y+1$, does not satisfy the initial condition.