 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1/17/a/solution

v = x y

. The product rule gives

v^{'} = x y^{'} + y

, turning the equation into

v^{'} = e^{v}

. Consequently

(e^{- v})^{'} = - 1

, so

e^{- v} = C - x

x = 1

the initial value gives

v = 1

, hence

C = 1 + e^{- 1}

. Recovering

y = v / x

gives

y (x) = - \frac{ln ( 1 + e ^{- 1} - x )}{x}, 0 < x < 1 + e^{- 1} .

This is the maximal real graph interval containing the initial point: the logarithm requires

x < 1 + e^{- 1}

, and the numerator has a nonzero limit at zero, so this particular solution cannot extend through zero. Differentiating

x y = - ln (C - x)

verifies

x y^{'} + y = 1/ (C - x) = e^{x y}

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