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The negative drift moves the outflow boundary layer to $x=1$. The first-order outer equation is $-(1+x)y_0^{\prime }+y_0=0$. It must satisfy the left boundary condition, giving $y_0=1+x$. This tends to two at the right endpoint.

Use $X=(1-x)/ϵ$. The leading inner equation is $Y_0^{\prime \prime }+2Y_0^{\prime }=0$, and impose $Y_0(0)=1$ and $Y_0\to 2$ on matching. Thus $Y_0=2-e^{-2X}$, and a leading uniform asymptotic approximation isFor higher accuracy, expand the drift as $2-ϵX$ in the inner equation, solve the successive inhomogeneous equations, and match to outer corrections satisfying zero left endpoint data. There is no need for a left endpoint layer. Choosing the growing exponential at the wrong endpoint would prevent matching and is the reason the layer location matters.

The positive drift places the boundary layer at the left endpoint. Set $y_{\mathrm{out}}=y_0+ϵy_1+\cdots$ and impose the right boundary condition order by order. The leading ordinary differential equation is $[(1+x)y_0{]}^{\prime }=0$, so $y_0=2/(1+x)$. At the next order,which integrates toThe outer limit at the left endpoint is two, so it cannot satisfy the left boundary value without a rapidly varying correction.

For the inner expansion use $X=x/ϵ$ and $y=Y_0(X)+ϵY_1(X)+\cdots$. Its equations and boundary values areMatching the leading constant gives $Y_0=2-e^{-X}$. The first-order forcing is then $-2+(1-X)e^{-X}$. A particular solution is $-2X+\frac{1}{2}{X}^2{e}^{-X}$, and matching determines the constant to be $3/2$. Enforcing $Y_1(0)=0$ givesIn the overlap region the common expansion is $2+ϵ(3/2-2X)$. Adding the inner and outer approximations and subtracting that common part gives the uniform asymptotic approximationIt has error $O({ϵ}^2)$ uniformly for $0\le x\le 1$, apart from an exponentially small endpoint adjustment of the same asymptotic irrelevance. At $x=0$ it equals one exactly, and at $x=1$ it differs from one only exponentially. The polynomial multiplying the exponential remains bounded on the inner scale.

There is an independent check from a first integral. The equation is $[ϵy^{\prime }+(1+x)y{]}^{\prime }=0$, so its exact solution has the formwhere $C$ is selected by the right boundary value. Endpoint expansion of this integral gives $C=2-ϵ/2+O({ϵ}^2)$, the same outer coefficients and the same inner expansion. This check also establishes the uniform order of the composite approximation without treating an inner residual as an outer estimate.