What is the harmonic conjugate of $u=4xy-3x+5y$? I got $u'x=4-y=v'y$ then I integrated $v'y$ to get $v= 2y^2-3y+h(x)$.
Then I did $-u'y=v'x$ so, $5= h'(x)$ then I integrated $5$ with respect to $x$ to get $5x$ and so, my harmonic conjugate should be $v=5x+c$.
Where did I go wrong?
I didn't understand $100\%$ of what you done there, but it seems to me that you differentiated something wrong in the beginning.
You want $v(x,y)$ such that $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x},$$that is: $$\frac{\partial v}{\partial y}(x,y) = 4y-3, \quad \frac{\partial v}{\partial x}(x,y) = -4x-5.$$From the first one, we get $v(x,y) = 2y^2-3y + h(x)$. Then $\frac{\partial v}{\partial x}(x,y) = h'(x) = -4x-5$, whence $h(x) = -2x^2-5x+c$. We conclude that: $$v(x,y) = -2x^2 + 2y^2-5x -3y + c, \quad c \in \Bbb R.$$One readily checks that $\nabla^2v = 0$, as desired.