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Solve dy/dx = (x - y)² subject to y(1) = 1. The solution is:

1. - logₑ |(1 - x + y)/(1 + x - y)| = 2(x - 1)
2. logₑ |(2 - y)/(2 - x)| = 2(y - 1)
3. logₑ |(2 - x)/(2 - y)| = x - y
4. - logₑ |(1 + x - y)/(1 - x + y)| = x + y - 2

Correct answer: - logₑ |(1 - x + y)/(1 + x - y)| = 2(x - 1)

Solution

With v = x - y, dv/dx = 1 - v² gives (1/2) ln|(1+v)/(1-v)| = x + C; using v=0 at x=1 yields C=-1, which rearranges to -ln|(1 - x + y)/(1 + x - y)| = 2(x - 1).

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