 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/20/b/solution

With

u = X Y

, divide the separated equation on

x \neq = 0

y \neq = 0

to obtain

\frac{e ^{x} X ^{'}}{x X} = \frac{y Y ^{'}}{Y} = λ .

Thus

X^{'} / X = λ x e^{- x}

and

Y^{'} / Y = λ / y

X = A e^{- λ (x + 1) e^{- x}}, Y = B ∣ y ∣^{λ} .

Choose

λ = 1

and the signed branch

Y = y

. A nonconstant solution, smooth for every real

x, y

, is

u (x, y) = y exp [- (x + 1) e^{- x}] .

Indeed

u_{x} = x e^{- x} u

and

u_{y} = exp [- (x + 1) e^{- x}]

, so

e^{x} u_{x} - x y u_{y} = 0

. This verification includes the axes excluded temporarily during separation.

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