Question

If f(x,y)is a function satisfying euler’ s theorem then?

(a) \(x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}=n(n-1)f\)

(b) \(\frac{1}{x}^2 \frac{∂^2 f}{∂x^2}+2/xy \frac{∂^2 f}{∂x∂y}+\frac{1}{y}^2 \frac{∂^2 f}{∂y^2}=n(n-1)f\)

(c) \(x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}=nf\)

(d) \(y^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+x^2 \frac{∂^2 f}{∂y^2}=n(n-1)f\)

The question was posed to me at a job interview.

This is a very interesting question from Euler’s Theorem topic in division Partial Differentiation of Engineering Mathematics

Answer 1

Right choice is (a) \(x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}=n(n-1)f\)

Easiest explanation: Since f satisfies euler’s theorem,

\(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=nz\)

Differentiating it w.r.t x and y respectively we get,

\(x \frac{∂^2 u}{∂x^2}+\frac{∂u}{∂x}+y \frac{∂^2 u}{∂x∂y}=n \frac{∂u}{∂x}\),

and

\(x \frac{∂^2 u}{∂y}∂x+\frac{∂u}{∂y}+y \frac{∂^2 u}{∂y^2}=n \frac{∂u}{∂y}\)

Multiplying with x and y respectively,

\(x^2  \frac{∂^2 u}{∂x^2}+x \frac{∂u}{∂x}+xy \frac{∂^2 u}{∂x∂y}=nx \frac{∂u}{∂x}\),

and

\(xy \frac{∂^2 u}{∂y}∂x+y \frac{∂u}{∂y}+y^2 \frac{∂^2 u}{∂y^2}=ny \frac{∂u}{∂y}\)

Adding above equations we get

\(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}=n(n-1)u\)