Those are three different differential equations, not one.

The general solution to [tex]\frac{\partial^2u}{\partial x^2}= 0[/itex]isu(x,y)= Ax+ g(y). But that does not necessarily satisfy the other equations. If you put u(x,y)= Ax+ g(y) into the third equation, you get [tex]\frac{d^2g}{dy^2}= 0[tex] which has general solution g(y)= By+ C. So u(x,y)= Ax+ By+ C. You can then check to see that it also satisfies the second equation: .