 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/8/b/iv/solution

Write

t = 1 - x

. The analytic solution has

y_{1} = a_{0} + O (t)

with

a_{0} \neq = 0

\frac{1}{( 1 - x ) y _{1} ( x ) ^{2}} = \frac{1}{a _{0}^{2} ( 1 - x )} + O (1) .

Integrating with respect to

x

gives

- a_{0}^{- 2} lo g (1 - x) + O (1)

y_{1}

therefore gives

y_{2} (x) = - \frac{1}{a _{0}} lo g (1 - x) + O (1) .

Any solution independent of

y_{1}

has a nonzero coefficient of this second solution. Hence

y (x) = K lo g (1 - x) + O (1), K \neq = 0, x \to 1^{-} .

An added multiple of

y_{1}

stays bounded and cannot cancel this logarithm. On the real side

x > 1

, the corresponding statement uses

lo g ∣1 - x ∣

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