Question

Let u = `log (x^4 - y^4)/(x - y).` Using Euler’s theorem show that `x (del"u")/(del"x") + y(del"u")/(del"y")` = 3.

Answer 1

Given u = `log (x^4 - y^4)/(x - y)`

Taking exponential both sides,

`e^"u" = ((x^4 - y^4)/(x - y))   ....[because e^(log x) = x]`

Let f = `"e"^"u" = (x^4 - y^4)/(x - y)`

∴ f(tx, ty) = `(("t"x)^4 - ("t"y)^4)/(tx - ty)`

`= ("t"^4 (x^4 - y^4))/("t" (x - y))`

`= "t"^3 ((x^4 - y^4)/(x - y))`

= t³f (x, y)

∴ f is a homogeneous function of degree 3.

By Euler’s theorem,

`x (del"f")/(del"x") + y(del"f")/(del"y")` = nf

`=> x * (del"f")/(del"x") + y * (del"f")/(del"y")` = 3f

`=> x * (del)/(delx) (e^"u") + y * (del)/(del "y") (e^"u") = 3 * "e"^"u"    ...[because "f" = e^"u']`

`=> x * e^"u" (del^"u")/(del x) + y * e^"u" (del "u")/(del "y") = 3e^"u"`

Dividing throughout by e^u, we get

`x (del "u")/(del x) + "y" (del "u")/(del "y")` = 3

Hence proved.