Question

If \(u = Tan^{-1} (\frac{x^3+y^3}{x+y})\) then, \(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}\) is?

(a) Sin(4u) – Cos(2u)

(b) Sin(4u) – Sin(2u)

(c) Cos(4u) – Sin(2u)

(d) Cos(4u) – Cos(2u)

The question was posed to me during an online exam.

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

Answer 1

The correct answer is (b) Sin(4u) – Sin(2u)

Explanation: Let, v = Tan(u) = x^2 f(y/x)

By euler’s theorem,

g(u) = \(x \frac{∂u}{∂x}+y \frac{∂u}{∂y} = 2\frac{Tan(u)}{Sec^2 (u)} = Sin(2u)\)

Hence,

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

\(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}\) = Sin(2u)[2Cos(2u)-1] = Sin(4u)-Sin(2u)