STEP III 1997 question 6 solution

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Solution 1

If then .

So

So as desired.

This has general solution .

.

To use the 2nd boundary condition, observe and so we have:

and so .

So (which is also ). Finally, we need to show:

.

But , hence result.

Solution by DFranklin

Solution 2

(1)

Differentiate (1) to get (2):

Substitute (1), (2) and into to get

The general solution of this linear homogeneous second order differential equation is We are given y(1)=1, hence after the tranformation of above, we have the boundary condition y(0)=1 which results in A=1. Furthermore, we are given which really means that is an even function when n is either 0 or even. And that it's an odd function when n is odd. But after the substitution of , becomes an even function for all n since cosine is an even function.

We have which can only be even for all n when B=0

and for

It follows that and as required.

Upon expanding with Trigonometric Addition Formulas and adding and :

as shown.

Solution by khaixiang