Given function:
$u=x^3-3xy^2$.
I have solved this problem in two different methods, but two methods give two different answer. Why?
Here these are:
$u_x=3x^2-3y^2, u_y=-6xy\\
u_{xx}=6x, u_{yy}=-6x$
So, $\nabla^2 u=0\implies u$ is harmonic function.
Let $v$ be conjugate harmonic function. Then it satisfies C-R equations:
$u_x=v_y$ and $u_y=-v_x$
method 1:
$dv=v_x dx+ v_y dy\\ =-u_y dx+ u_x dy\\ =6xy dx+ (3x^2-3y^2) dy\\ v=3x^2y+(3x^2y-y^3)+c\\ v=6x^2y-y^3+c$
method 2:
$u_x=3x^2-3y^2\\
\implies v_y=3x^2-3y^2\\
\implies v=3x^2y-y^3+\phi(x)\\
\implies v_x=6xy+\phi'(x)\\
\implies -u_y= 6xy+\phi'(x)\\
\implies 6xy= 6xy+\phi'(x)\\
\implies \phi'(x)=0\\
\implies \phi(x)=c $.
So, $v= 3x^2y-y^3+c$.
Above two methods give two different functions $v$, why? But two methods should give same $v$. Please help