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Assume $ϵ\to 0^{+}$. Since the drift $(1+x{)}^2$ is positive, the fast mode decays to the right; the matched asymptotic expansion has a left boundary layer at $x=0$ of width $O(ϵ)$, and the outer expansion uses the condition at $x=1$.

Write $y\sim y_0+ϵy_1$. The first two equations for the outer expansion areEnforcing $y_0(1)=1$ and $y_1(1)=0$, defineThen the outer asymptotic expansion isTo derive the correction, set $y_1=Eq$. Cancellation of the homogeneous terms leaves $q^{\prime }=-2/(1+x{)}^5-1/(1+x{)}^6$; integration and $q(1)=0$ give the expression above.

In the left boundary layer, set $\xi =x/ϵ$ and $y=Y_0(\xi )+ϵY_1(\xi )+\cdots$. The equation becomes $Y^{\prime \prime }+(1+ϵ\xi {)}^2{Y}^{\prime }+ϵY=0$. Put $A=e^{1/2}$ and $B=1-A$. The leading inner equation and matching give $Y_0^{\prime \prime }+Y_0^{\prime }=0$, $Y_0(0)=1$, $Y_0(\infty )=A$, hence $Y_0=A+B{e}^{-\xi }$. At the next order,Since $q(0)=53/80$, put $C=53A/80$. The matched inner correction satisfying $Y_1(0)=0$ isFor verification, the differential operator $d^2/d{\xi }^2+d/d\xi$ maps $e^{-\xi }P(\xi )$ to $e^{-\xi }(P^{\prime \prime }-P^{\prime })$. The large-$\xi$ common part is $A+ϵ(C-A\xi )$, precisely the small-$x$ outer expansion in the overlap $ϵ\ll x\ll 1$.

Adding the inner expansions and outer expansions and subtracting their common part gives the first-order composite expansion for a positive driftThis is uniformly correct through $O(ϵ)$ on $0\le x\le 1$. It satisfies the left boundary condition exactly; at the right boundary the remaining error is exponentially small. Polynomial growth in the inner correction is harmless because it is multiplied by the decaying exponential. For the smooth nonvanishing drift, the next matched terms are uniformly $O({ϵ}^2)$.

For fixed $x>0$, set $y=y_0+\epsilon y_1+{\epsilon }^2{y}_2+\cdots$ and impose the data at one at each order. The leading equation gives $y_0^{\prime }=0$, hence $y_0=1$. At the next two orders,The boundary conditions $y_1(1)=y_2(1)=0$ give the three-term outer expansionIts ordering fails when $\epsilon /x=O(1)$, locating a boundary layer of thickness $O(\epsilon )$ at zero. Put $x=\epsilon \xi$ and $y=Y_0(\xi )+\epsilon Y_1(\xi )+\cdots$. The rescaled equation isIts leading equation is ${\xi }^2{Y}_0^{\prime }=Y_0^3$. Matching to the positive outer value one fixes the sign and integration constant:At the next order the bracket $\xi (Y_0^2-1)+2Y_0^2$ vanishes identically, leavingThe general solution is $Y_1=(1+k\xi ){\xi }^{1/2}/(\xi +2{)}^{3/2}$. In the overlap its expansion isThe outer solution re-expressed on this scale has the corresponding terms $1-1/\xi +3/(2{\xi }^2)+\epsilon (1-2/\xi )+\cdots$. Thus matched asymptotic expansion determines $k=1$, consistently in both displayed orders, and the two-term inner expansion isThis positive branch tends continuously to zero at $x=0$, with a square-root cusp. An infinite endpoint derivative is compatible with the degeneracy of the original equation's $x^2{y}^{\prime }$ coefficient.

At fixed positive $x$ the leading outer expansion satisfies $z_0^{\prime }+z_0=e^{-x}/x$. Integrating and imposing the value at one yieldsIt diverges negatively at zero, and the neglected nonlinear shift in the derivative coefficient becomes important when $x\sim \epsilon |\log x|$. Thus the relevant logarithmically enhanced nonlinear boundary layer is larger than a plain $O(\epsilon )$ layer. Write $L=\log (1/\epsilon )$, $x=\epsilon LX$ andFor fixed $X$, the leading inner equation is $(X+1)Z_X=1$. Matching to the outer logarithm, for which $Z\sim \log X$, sets the integration constant and givesIn particular the exact equation at zero is $-\epsilon z(0)z^{\prime }(0)=1$, so

An implicit inner form also checks the matching constants. With $x=\epsilon \xi$, retain $z$ without assigning it a bounded size. The leading equation is $(\xi -z)z_{\xi }=1$. Inverting it gives $d\xi /dz-\xi =-z$, henceIts large-$\xi$ match to $z\sim 1+\log \epsilon +\log \xi$ fixes $B\sim e^{-1}/\epsilon$. Taking that matched leading value at zero giveswhere $W_0$ is the positive real branch of the Lambert W function. Its large-argument expansion reproduces the logarithm and log-log terms above. This refinement is a matched approximation, not an exact solution of the full equation; the leading slope conclusion follows directly from the first inner scaling and the exact endpoint identity.

Define the nested integralThe order-$s^2$ equation in the inner expansion is ${\mathcal{L}}_0{Y}_2=2z{Y}_1$. The preceding result decomposes its forcing into $2z^{2}E+2bzE-2bz+2bzg$. The particular solutions combine toFor example, ${\mathcal{L}}_{0}H=E$ follows directly by differentiating the nested integral, so the equation can also be verified without relying on the supplied particular-solution formulas. At zero $Y_{2,p}=b^2$; add $-b^2+DE$ to impose the left boundary condition. As $z\to \infty$ this givesThe outer expansion in the overlap, derived in the previous section, requires the remaining constant to be $-\frac{1}{2}\log s$. Thus $D=C+b^2-\frac{1}{2}\log s$. The logarithmic matching constant is an error-function logarithmic switchback.

The complete answer through order $\epsilon$ in the inner expansion isHere $E=\mathrm{erf}z$, $g=e^{-z^2}$ and $b=2/\sqrt{\pi }$. This expression is zero at $z=0$, and its large-$z$ expansion agrees with $1+x+x^2/2-(\epsilon /2)\log x$. Together with $y_{\mathrm{out}}=e^x[1-(\epsilon /2)\log x]$, it covers both requested regions. The overlap uses $1\ll z\ll {\epsilon }^{-1/2}$, with the usual termwise ordering restrictions when matching higher terms.

If a single continuous approximation is desired, an additive composite expansion isThe apparent logarithmic divergence at $x=0$ cancels: its residual is $-(\epsilon /2)(e^x-1)\log x\to 0$. The composite therefore extends continuously to the left endpoint and satisfies the right boundary condition to the requested order.

Write $E=\mathrm{erf}z$, $g=e^{-z^2}$ and $b=2/\sqrt{\pi }$. The first correction in the inner expansion satisfies ${\mathcal{L}}_0{Y}_1=2zE$. Direct differentiation verifies that $zE+bg$ is a particular solution. Adding the constant and error function homogeneous solutions gives $Y_1=zE+bg+A+BE$.

The left boundary condition forces $A=-b$. Re-expanding the outer expansion at $x=sz$ givesTherefore $Y_1\sim z$, with no constant term in its large-$z$ overlap. Since $E\to 1$ and $g\to 0$, this sets $B=b$. Thus the required half-order correction isIt vanishes at zero and has the correct linear overlap. Keeping this $O(\sqrt{\epsilon })$ term is essential even though the outer solution has no correction of that order.

For the matched asymptotic expansion, first determine the outer expansion. Setting $\epsilon =0$ away from $x=0$ gives $y_0^{\prime }=y_0$, and the right boundary condition sets $y_0=e^x$. At first order,soThere is no outer $\sqrt{\epsilon }$ term: its homogeneous coefficient would be forced to zero by the right boundary condition. The logarithmic overlap forces a switchback term in the inner expansion.

Near zero, balance $\epsilon y^{\prime \prime }$ against $2x{y}^{\prime }$ to get the square-root boundary layer at a vanishing drift:For fixed $z$ expand $Y=Y_0+s{Y}_1+s^2{Y}_2+\cdots$, allowing logarithmic dependence on $s$ in $Y_2$. With ${\mathcal{L}}_0={\partial }_z^2+2z{\partial }_z$, the leading equation is ${\mathcal{L}}_0{Y}_0=0$. Its bounded solutions are a constant plus an error function. The left boundary condition and the overlap value one fixThe following two solution sections complete the half-order and first-order matching; their headings retain the scaffold's organization of the supplied hints.

For the outer expansion, the first-order reduced ordinary differential equation gives $w_0=e^{-x^2}$ after imposing the right boundary condition. Its first correction satisfiesAn integrating factor givesThe left boundary is at $x=\epsilon$, where the highest-derivative coefficient is already ${\epsilon }^2$. Balancing that coefficient against $w^{\prime }$ gives a layer of width ${\epsilon }^2$, not $\epsilon$. This is a boundary layer around a moving endpoint:At leading order $W_0^{\prime \prime }+W_0^{\prime }=0$, and matching to the outer limit one while setting $W_0(0)=0$ gives $W_0=1-e^{-z}$. The first correction satisfiesDirect differentiation of $-z(z+2)e^{-z}/2$ verifies the forcing. It vanishes both at zero and at infinity, so no homogeneous term is needed. The requested inner expansion isIts order-$\epsilon$ overlap constant is zero, agreeing with the outer expansion at the moving endpoint. The overlap can be chosen with $1\ll z\ll {\epsilon }^{-1}$.

An independent scaling check drops the reaction term only in the layer and integrates $\epsilon x{w}^{\prime \prime }+w^{\prime }=0$. Its decaying fast solution is proportional to $(x/\epsilon {)}^{1-1/\epsilon }$. Sinceit gives exactly the same first correction. A useful additive composite expansion, with the left boundary value made exact, isThe exponentially small remainder at the right endpoint is beyond all retained algebraic orders.

The three distinguished limits are the outer expansion at $x=O(1)$, an intermediate asymptotic region at $x=O(ϵ)$, and an inner expansion at $x=O({ϵ}^2)$. The middle scale resolves the coefficient change from $x$ to $x+ϵ$; the narrower scale resolves the unmet left boundary condition. The original equation is understood on $x>0$ with a continuous extension to the boundary. As explained below, imposing a classical second derivative at $x=0$ would be too strong for this degenerate ordinary differential equation.

In the outer expansion, set $y=y_0+ϵy_1+\cdots$. The leading equation is $x(x+1)y_0^{\prime }-x{y}_0=0$. Enforcing the right boundary condition gives $y_0=x+1$. Because $y_0^{\prime \prime }=0$, the next equation is particularly simple:An integrating factor, or the substitution $y_1=(x+1)u$, gives $u^{\prime }=-1/[x(x+1)]$. ThusThe error statement is for $x$ bounded away from zero. The left limiting form is $1+x-ϵ\log (2x)$ through the terms needed for matching. A logarithm in the correction makes a pure integer-power inner ansatz inadequate.

For the intermediate asymptotic region, put $x=ϵs$ and $y=V(s)$. The transformed equation isThe leading $V_0$ is a constant fixed to $1$ by matching. There is no order-$\sqrt{ϵ}$ forcing, so that homogeneous constant is zero after matching. At order $ϵ$, $(s+1)V_1^{\prime }=s$, giving $V_1=s-\log (1+s)+C(ϵ)$. Matching for $1\ll s\ll 1/ϵ$ with the outer expansion fixes $C=-\log (2ϵ)$. HenceThe switchback term $-ϵ\log ϵ$ is larger than an ordinary $O(ϵ)$ constant correction and must be retained. The highest derivative term first changes this middle solution at order ${ϵ}^{3/2}$, so it does not alter the displayed result.

To locate the inner expansion, compare the derivative terms for $x\ll ϵ$. Their ratio at a layer width $\delta$ is $ϵ/\sqrt{\delta }$, so the square-root-degenerate endpoint layer has $\delta ={ϵ}^2$. Put $x={ϵ}^{2}X$ and $y=U(X)$. After dividing by the common leading factor, the equation through order $ϵ$ isBoth orders therefore have the same homogeneous operator $\sqrt{X}{d}^2/d{X}^2+d/dX$. Its derivative mode is proportional to $e^{-2\sqrt{X}}$. Using $U(0)=0$ and the supplied elementary integral, defineThe intermediate asymptotic region gives the overlap value $1-ϵ\log (2ϵ)$ as $s\to 0$. ThereforeIt satisfies the left boundary condition exactly to the retained orders, and its large-$X$ limit matches the middle result in $1\ll X\ll 1/ϵ$. The outer and middle limits match with $ϵ\ll x\ll 1$. These three formulae include all terms through $O(ϵ)$, including the logarithmic switchback; the omitted terms in the middle and inner regions need not be $O({ϵ}^2)$.

For a check on that last point, continue the middle equation by one order. Its ${ϵ}^{3/2}$ term obeysThe integration constant is selected by the outer limit. Its value at $s=0$ is $\pi /2$, so the inner matching amplitude has a next correction $\pi {ϵ}^{3/2}/2$. This independently explains why an $o(ϵ)$ remainder is appropriate.

If a single additive composite expansion is useful, combine the three formulae and subtract their two common overlaps. Writing $A=1-ϵ\log (2ϵ)$ and interpreting $x\log x=0$ at zero givesIt is zero at $x=0$, gives $2+O({ϵ}^2)$ at $x=1$, and reproduces all three retained expansions. It does not assert an $O({ϵ}^2)$ error uniformly through the nested regions.

Finally, near zero $H(X)=2X-\frac{8}{3}{X}^{3/2}+O(X^2)$. Thus the solution has a finite nonzero first derivative but generally a second derivative diverging like $x^{-1/2}$. This is compatible with the equation for $x>0$: the vanishing coefficient of $y^{\prime \prime }$ multiplies that singular derivative to give a finite limit. If one instead required $y\in C^2[0,1]$ and the differential equation pointwise at $0$, the left condition would force $y^{\prime }(0)=0$. The normalized equation's other coefficients have only integrable $x^{-1/2}$ singularities, so uniqueness for the initial value problem with $y(0)=y^{\prime }(0)=0$ would give the zero solution, contradicting the right condition. Explicitly, write the normalized equation as a first-order system for $(y,y^{\prime })$ with an integrable coefficient matrix; its integral equation and the Gronwall inequality force a solution with zero initial vector to vanish. The continuous-endpoint interpretation is therefore essential.

The first of the Gaussian drift primitives, $E_0$, is an even function; the second, $E_1$, is an odd function. Their derivatives follow by the fundamental theorem of calculus. In particular $E_0^{\prime }$ is the Dawson function, which behaves as $1/(2z)+O(z^{-3})$. Since the error function tends to $\pm 1$, the derivative of $E_1$ behaves as $1/(2|z|)+O(|z{|}^{-3})$ at either end. Integration therefore givesAlso $F\sim z+\sqrt{\pi }/4$ on the left and $F\sim \sqrt{\pi }/4$ on the right. The two inner overlaps are consequentlyExpressing the outer expansions in $r=\epsilon z$ requires the left constant $1+A\log \epsilon$ and the right constant $B\log \epsilon$. Equating those constants gives the stated $\alpha ,\beta$. This is exactly how a logarithmic overlap creates a switchback term: $\log |r|=\log \epsilon +\log |z|$. Adding the outer solutions and removing these overlaps leaves the displayed additive composite expansion.

The solution and its first derivative are continuous at $x=1$: a jump would create a Dirac delta distribution or its derivative, absent from the forcing. Put $r=x-1$, $A=a+1$ and $B=b$. The outer expansions, fixed by the respective endpoint boundary conditions, areBalancing diffusion and advection near $r=0$ gives the interior layer at a simple zero of advection with inner variable $z=r/\epsilon$. The leading inner expansion solves $Y_0^{\prime \prime }+2z{Y}_0^{\prime }=0$, so matching to $A$ and $B$ givesFor the next term define the Gaussian drift primitivesWriting $H$ for the Heaviside step function, let $F(z)=zH(-z)+(\sqrt{\pi }/4)\mathrm{erf}|z|$. Its value and first derivative match at zero and $(D^2+2zD)F=2zH(-z)$. Thuswhere the logarithmic overlaps fixHere $C_1,C_2$ are the constants in the supplied large-positive-$z$ limits of $E_0,E_1$. The terms $\epsilon \log \epsilon$ are essential switchback terms; discarding them would not give accuracy through $O(\epsilon )$.

Subtracting the common overlap from the inner expansion and outer expansions gives the additive composite expansionThe standard overlap subtraction leaves $\epsilon rH(-r)$. Replacing $H(-r)$ by $(1-\mathrm{erf}z)/2$ changes the value only by $O({\epsilon }^2)$ uniformly and gives a composite with continuous first derivative. The last term restores the endpoint values through the retained order. When $b=a+1$, $d=0$ and the order-one error-function jump disappears. An $O(\epsilon )$ interior adjustment remains to accommodate the leading outer derivative mismatch, together with logarithmic matching when $A\ne 0$.

Reversing the diffusion sign changes the leading inner equation to $-Y_0^{\prime \prime }+2z{Y}_0^{\prime }=0$. Its nonconstant solution grows like the integral of $e^{z^2}$, so bounded matching forces the same leading value $C$ on both sides. The bulk is $C+r$ on the left and $C$ on the right; both endpoint conditions are instead supplied by decaying endpoint layers for reversed diffusion, of width ${\epsilon }^2$. A weaker width-$\epsilon$ interior adjustment matches derivatives and selects $C$. Indeed, its first-order equation has a non-growing solution only ifThe leading structure is consequentlywith the smaller interior correction understood. Unlike the positive-diffusion case, the endpoint values are not transported into two distinct order-one inner limits.

At fixed $r$, the leading equation is $(r^2{f}_0^{\prime }{)}^{\prime }=0$. Its solution compatible with the eventual decaying far field and the boundary value at one is $f_0=1/r$. This decay implies $f=O(ϵ)$ at distances $r=O({ϵ}^{-1})$. Balancing radial derivatives against the linear screening term identifiesThe distant region is governed by the modified Helmholtz equation; its decaying homogeneous profile is $e^{-x}/x$.

For a matched asymptotic expansion, write the fixed-$r$ approximation as $f=f_0+ϵf_1+{ϵ}^2{f}_2+\cdots$, allowing logarithms of $ϵ$ in the coefficients. Successive equations areThe boundary condition imposes $f_0(1)=1$ and $f_1(1)=f_2(1)=0$. Before matching, their integrated forms can be writtenIn particular a pure power series with parameter-independent coefficients will be insufficient: the logarithmic overlap creates a switchback term.

In the distant region set $f=ϵF_0(x)+{ϵ}^2{F}_1(x)+\cdots$. The scaled equation is $f_{xx}+2f_x/x-f=x{f}^3/ϵ$, soDecay and leading matching give $F_0=e^{-x}/x$. The radial modified Helmholtz equation gives the supplied particular integral in terms of the exponential integral:As $x\downarrow 0$, the small-argument expansion of the exponential integral yieldswhere $\gamma$ is the Euler--Mascheroni constant. Substitute $x=ϵr$ to compare the two expansions in $1\ll r\ll {ϵ}^{-1}$:The constant at order $ϵ$ fixes $C_1=-1$. Its $1/r$ coefficient then fixes $\alpha +\frac{1}{2}\log 2=1$, and the constant at order ${ϵ}^2$ fixes $D$. HenceThe required inner expansion at fixed $r$ isAt fixed positive $x=ϵr$, the outer expansion isBoth display every term through the requested order, including the ${ϵ}^2\log ϵ$ switchback term in the fixed-$r$ region.

To form an additive composite expansion, subtract the common overlap from the sum of the inner and outer expressions. Their retained common part isThus one composite is $ϵF_0(ϵr)+{ϵ}^2{F}_1(ϵr)-{ϵ}^2(D+1/2)/r$. The last term can be screened by multiplying it by $e^{-ϵr}$ without changing either retained expansion. This gives a useful exponentially decaying version:Its boundary value is $1+o({ϵ}^2)$. If exact satisfaction of the boundary value is desired, use insteadThe small-argument expansion of the exponential integral shows that this normalized composite has the same two retained expansions; it equals one at $r=1$ and tends to zero at infinity.

The reduced first-order equation cannot satisfy both endpoint values. The given outer expansion satisfies $y(1)=2$ but has $y_0(0)=e$, so the boundary layer lies at $x=0$ and has stretched coordinate $X=x/ϵ$.

Write the inner expansion as $Y=Y_0+ϵY_1+\cdots$. After multiplying the differential equation by $ϵ$, it becomesAt leading order,The boundary condition $Y_0(0)=0$ and matching to $y_0(0)=e$ giveAt the next order,Matching to $y_1(0)=e(\ln 2-1)$ and imposing $Y_1(0)=0$ gives

The additive composite expansion is outer plus inner minus their common part. With $X=x/ϵ$, it isIt satisfies $y_{\mathrm{comp}}(0)=0$ exactly through the retained order, satisfies the right boundary condition up to exponentially small terms, and is uniformly accurate to $O(ϵ)$.

Introduce the stretched coordinateThe equation becomesSeek the three-term outer expansionAt first order,so decay at infinity and matching giveAt the next order,Using the function $Y$ supplied in the question, the matched decaying solution isHere $Y$ is the stated particular solution normalized to decay at infinity.