We consider the differential equation

$ \epsilon y’’+ 2yy’ -4xy =0 $

With boundary condition $ y(0)=-1 $, $ y(1)=2$

How could we find the inner solution at $x$ nears $0$? Because if $y(0)$ changes a little, I could show that it is a boundary layer or interior layer; However at this exact $y(0)=-1$ I can't match with the outer solution successfully.

  • $\begingroup$ This question should help: math.stackexchange.com/q/1546644/131807 $\endgroup$ – David Feb 26 '18 at 23:57
  • $\begingroup$ You should perhaps indicate what you found as inner and outer solution and that the result should be something like $y(x)=\tanh((x-c(ϵ))/ϵ)+x^2$ and the fitting problem is to derive the form of $c(ϵ)$ from the equation. $\endgroup$ – LutzL Mar 6 '18 at 18:40

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