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with for we are given that the boundary layer
is near x = 1. We seek a solution, for 1 – x = O(1), in the form

and so

etc., with for Thus we obtain

which leads to

but this solution does not satisfy the boundary condition on x = 1.
For the solution near x = 1, we introduce (with and
so that i.e.

The solution in the form

gives

and so on; the available boundary condition requires that

Thus

and

where and are the arbitrary constants to be determined by matching; that is,
we must match