 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/4/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 4 / d / Solution by

Codex  0 2026-10-31

For

x \neq = \pm 1

, write the linear ordinary differential equation as

y^{'} + \frac{2 x}{1 - x ^{2}} y = \frac{x ( 1 - x )}{1 - x ^{2}} .

An integrating factor on each interval avoiding

\pm 1

is

μ (x) = 1/ (1 - x^{2})

; multiplication by a nonzero constant, including a sign, is immaterial. Therefore

(\frac{y}{1 - x ^{2}})^{'} = \frac{x ( 1 - x )}{( 1 - x ^{2} ) ^{2}} = \frac{x}{( 1 - x ) ( 1 + x ) ^{2}} .

The partial fraction decomposition is

\frac{x}{( 1 - x ) ( 1 + x ) ^{2}} = \frac{1}{4 ( 1 - x )} + \frac{1}{4 ( 1 + x )} - \frac{1}{2 ( 1 + x ) ^{2}} .

Integrating gives the general solution

y (x) = \frac{1 - x}{2} + (1 - x^{2}) (C + \frac{1}{4} lo g \frac{1 + x}{1 - x}), x \neq = \pm 1.

The constant is independent on each connected interval of definition. At the singular points the undivided differential equation forces

y (1) = 0

and

y (- 1) = 1

. The displayed family tends to these values, but its derivative has a logarithmic divergence there, so no member extends as a differentiable function through either singular point.

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