# A question about perturbation method of

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.

• This question should help: math.stackexchange.com/q/1546644/131807 – David Feb 26 '18 at 23:57
• 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. – LutzL Mar 6 '18 at 18:40