 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016/9/b/solution

The initial point lies in

x > 1

, where division by

ln x

y^{'} + \frac{2}{x l n x} y = \frac{1}{l n x}

(ln x)^{2}

. Hence

\frac{d}{d x} [y (ln x)^{2}] = ln x, y (ln x)^{2} = x ln x - x + C .

The condition at

x = e

C = 0

, so

y (x) = \frac{x ( ln x - 1 )}{( ln x ) ^{2}}, x > 1.

The solution cannot be continued regularly through

x = 1

: its numerator tends to

- 1

while the denominator tends to zero. The real logarithm also requires

x > 0

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