Question

Verify Euler’s theorem for the function u = x³ + y³ + 3xy².

Answer 1

u = x³ + y³ + 3xy²

i.e., u(x, y) = x³ + y³ + 3xy²

u(tx, ty) = (tx)³ + (ty)³ + 3(tx) (ty)²

= t³x³ + t³y³ + 3tx (t²y²)

= t³(x³ + y³ + 3xy²)

= t³u

∴ u is a homogeneous function in x and y of degree 3.

∴ By Euler’s theorem, `x * (del"u")/(delx) + y * (del"u")/(dely)` = 3u

Verification:

u = x³ + y³ + 3xy² 

`(del"u")/(delx) = 3x^2 + 0 + 3y^2 delk/(delx)`(x)

= 3x² + 3y²(1)

= 3x² + 3y² …….. (1)

`therefore x * (del"u")/(delx)` = 3x³ + 3xy² 

`(del"u")/(dely) = 0 + 3y^2 + 3x(2y) = 3y^2 + 6xy`

`y * (del"u")/(dely)` = 3y³ + 6xy² ……… (2)

∴ (1) + (2) gives

`x * (del"u")/(delx) + y * (del"u")/(dely)` = 3x³ + 3y³ + 9xy² 

= 3(x³ + y³ + 3xy²)

= 3u

Hence Euler’s theorem is verified.