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Part BCSIR NET December 2024the-euler-operator-means-homogeneous

The euler operator means homogeneous

Let u = u(x, y) be the solution of the Cauchy problem x∂u/∂x+y∂u/∂y=u,(x,y)≠(0,0)x \partial{}u/\partial{}x + y \partial{}u/\partial{}y = u, (x, y) \ne (0, 0), with u(x,1)=1+x2,x∈Ru(x, 1) = \sqrt{1 + x^{2}}, x \in \mathbb{R}. Then which of the following statements is true?

  1. A.u(1, 0) = 0
  2. B.u(x1,y1)=u(x2,y2)u(x_{1}, y_{1}) = u(x_{2}, y_{2}) whenever x12+y12=x22+y22x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2}✓ Correct
  3. C.u(1,y)=2u(1, y) = \sqrt{2} for all y∈Ry \in \mathbb{R}
  4. D.u(x1,y1)=u(x2,y2)u(x_{1}, y_{1}) = u(x_{2}, y_{2}) whenever x1+y1=x2+y2x_{1} + y_{1} = x_{2} + y_{2}

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests standard counterexample.

See pricing

50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

Related counterexample: A first-order quasilinear Cauchy problem with smooth data has a global smooth solution

More on this topic

The chapter behind this: First-order PDE: characteristics, Lagrange and Charpit — free to read

From Partial Differential Equations › First-order PDE: Lagrange, Charpit, characteristics

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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