 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2015/5/a/ii/solution

- x^{2} (d y)^{2} + (d x)^{2} = 0, \frac{d y}{d x} = \pm \frac{1}{x} .

ξ = y + ln ∣ x ∣, η = y - ln ∣ x ∣.

2/ x

, so this is an invertible local change of variables whenever

x \neq = 0

. Write

u (x, y) = U (ξ, η)

. The chain rule gives

u_{x} x^{2} u_{xx} u_{yy} = \frac{U _{ξ} - U _{η}}{x}, = U_{ξξ} - 2 U_{ξ η} + U_{ηη} - U_{ξ} + U_{η}, = U_{ξξ} + 2 U_{ξ η} + U_{ηη} .

Substitution cancels both unmixed second derivatives and both first derivatives:

- x^{2} u_{xx} + u_{yy} - x u_{x} = 4 U_{ξ η} .

U_{ξ η} = 1.

The cancellation of the first derivatives is the key feature of these logarithmic characteristic coordinates.

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