Question

If \(u = e^{\frac{(x^2+y^2)}{x+y}}\) Then, \(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}\)=?

(a) u ln⁡(u)

(b) u ln⁡(u)^2

(c) u [1+ln⁡(u)]

(d) 0

I had been asked this question at a job interview.

I want to ask this question from Euler’s Theorem in chapter Partial Differentiation of Engineering Mathematics

Answer 1

The correct option is (b) u ln⁡(u)^2

The best I can explain: Let, v = ln(u) = \(\frac{x^2+y^2}{x+y} = x f(\frac{y}{x})\)

Hence by applying euler theorem,

\(x \frac{∂v}{∂x}+y \frac{∂v}{∂y}=v\)

Hence,

g(u) = \(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}=u ln⁡(u)\)

Hence,

\(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}\) = g(u)[g’(u)-1]

\(x^2  \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}\) = u ln(u)[1+ln⁡(u)-1] = u ln⁡(u)^2