Question

If `u=x^2tan^(-1)(y/x)-y^2tan^(-1)(x/y)`,prove that `(del^2u)/(delxdely)=(x^2-y^2)/(x^2+y^2)`

Answer 1

`u = x^2 tan^-1(y/x)-y^2tan^-1(x/y)`
`=> (del u)/ (del x) = 2x tan^-1(y/x) + x^2(x^2/(x^2+y^2))(-y/x^2) - y^2(y^2/(x^2+y^2))(1/y) `
`=> (del u)/ (del x) = 2x tan^-1(y/x) - (x^2y)/(x^2+y^2) - (y^3)/(x^2+y^2)`
Now, `(del^2u)/(delxdely) = 2x(x^2/(x^2+y^2))(1/x) - [(x^2(x^2+y^2)-x^2y(2y))/(x^2+y^2)^2] - [(3y^2(x^2+y^2)- y^3(2y))/(x^2+y^2)^2]`
`(del^2u)/(delxdely) = (2x^2)/(x^2+y^2) - [(x^4+x^2y^2-2x^2y^2+3x^2y^2+3y^4-2y^4)/(x^2+y^2)^2]`
`(del^2u)/(delxdely) = (2x^2)/(x^2+y^2) - [(x^4+2x^2y^2+y^4)/(x^2+y^2)^2]`
`(del^2u)/(delxdely) = (2x^2)/(x^2+y^2) - [(x^2+y^2)^2/(x^2+y^2)^2]`
`(del^2u)/(delxdely) = (2x^2)/(x^2+y^2) -1`
`(del^2u)/(delxdely) = (x^2 - y^2)/(x^2+y^2)`