 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/13/a/solution

Let

M = 2 x + e^{y}

and

N = x e^{y} - cos y

. Their cross partial derivatives agree:

M_{y} = e^{y} = N_{x}

. On the simply connected plane, this makes the differential form exact. Integrating

F_{x} = M

gives

F = x^{2} + x e^{y} + h (y)

; matching

F_{y} = N

then requires

h^{'} (y) = - cos y

, so one potential is

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

d F / d x = 0

along a solution. Its initial condition fixes the constant to

F (1, π /2) = e^{π /2}

. Hence the solution is specified implicitly by

x^{2} + x e^{y} - sin y = e^{π /2}, y (1) = \frac{π}{2} .

At the initial point

F_{y} = e^{π /2} \neq = 0

, so the implicit function theorem guarantees a unique local branch

y (x)

through that point.

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