 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/11/b/solution

First take

a > 0

and

∣ x ∣ < a

, and put

u = arcsin (x / a)

q = a^{2} - x^{2}

. Then

u^{'} = 1/ q

. The chain rule gives

y^{'} = \frac{2 u}{q}, y^{''} = \frac{2}{q ^{2}} + \frac{2 xu}{q ^{3}} .

Consequently

(a^{2} - x^{2}) y^{''} - x y^{'} = q^{2} (\frac{2}{q ^{2}} + \frac{2 xu}{q ^{3}}) - \frac{2 xu}{q} = 2,

which proves the identity on the open domain of differentiability. The squared inverse sine is unchanged when

a

is replaced by

- a

, so the identity also holds for negative

a

with

∣ x ∣ < ∣ a ∣

. The value

a = 0

is excluded.

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