Advanced Ordinary Differential Equation

Consider the boundary value problem

Ey"+y'=2t, y(0) =y(1)=1

For the function y=y(t), where E is a small parameter, 0<E<<1. You are given that the function y has a boundary layer close to the point t=0.

Q, Find the lowest order term, yo(t), in the outer expansion:

y(t) = yo(t) = Ey1(t) + ... (*)

A, Ok here's my workings...

Subbing (*) into boundary value problem i get:

E(yo" + Ey1" + ...) + (yo'+Ey1' + ...) = 2t

y(0) = yo(0) + Ey1(0) + ... = 1

y(1) = yo(1) + Ey1(1) + ... = 1

Therefore equating coefficients:

E^0: yo' = 2t, yo(0) = yo(1) = 1

Integrating yo' = 2t gives me: yo = t^2 + C

So then to work out the constant I then apply the initial conditions either, either y(0) = 1 or y(1)=1?, but obviously this would give me two different values for c? In my solutions they use y(1) = 1 which makes C = 0, so that yo = t^2. Could anyone shed any light upon which this is though, would be much appreciated.

Thanks.