Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 68 / 3 / b / Solution

Seek an outer asymptotic expansion $y=y_0+ϵy_1+\cdots$ for fixed $x>0$. The leading equation is $x{y}_0^{\prime }=0$, and the endpoint condition gives $y_0=1$. At first order, $x{y}_1^{\prime }=y_0=1$ and $y_1(1)=0$, so $y_1=\log x$. This outer correction cannot be evaluated at $x=0$.

The coefficient $x+ϵ$ selects the boundary layer coordinate $X=x/ϵ$. With $Y(X)=y(ϵX)$, the equation becomes $(1+X)Y_X=ϵY$. Its inner leading term is constant; matching it to the outer solution gives $Y_0=1$. At first order, $Y_{1X}=1/(1+X)$, hence $Y_1=\log (1+X)+C$. In the overlap $1\ll X\ll {ϵ}^{-1}$, the outer solution is $1+ϵ\log ϵ+ϵ\log X$. Matching the constants therefore givesfor fixed $X$. The logarithmic matching in a small-exponent boundary layer must retain $ϵ\log ϵ$, which is larger in magnitude than a plain order-$ϵ$ term. At the endpoint,The coefficient of a separate plain $ϵ$ term is zero.

Direct integration provides the exact check:At zero its logarithm is $ϵ\log ϵ-ϵ\log (1+ϵ)=ϵ\log ϵ-{ϵ}^2+O({ϵ}^3)$. Exponentiating gives $y(0)=1+ϵ\log ϵ+O({ϵ}^2{\log }^2ϵ)$, whose remainder is indeed $o(ϵ)$. The first-order uniform composite $1+ϵ\log [(x+ϵ)/(1+ϵ)]$ also satisfies the boundary condition exactly and reproduces both expansions to the requested order.