 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1/16/b/solution

F (x, y) = y e^{x^{2}} - x^{2}

. Its differential is

d F = 2 x (y e^{x^{2}} - 1) d x + e^{x^{2}} d y,

which proves exactness directly. Multiplying the differential equation by

e^{x^{2}}

shows that a solution satisfies

d F / d x = 0

, so

y e^{x^{2}} - x^{2} = C

. The initial value at zero gives

C = - 1

. Therefore

y (x) = (x^{2} - 1) e^{- x^{2}}, x \in R .

The factor

e^{x^{2}}

is also an integrating factor for the associated first-order linear equation.

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