Solving method for basic PDEs

How is it called the method used in the second and in the fourth of the following steps? I don't understand it that well. Usually, everything you do on a term of the equation you must do it on the other too. For example, in the second step there is an integration on the left side but only a "partial" integration on the right side. I don't understand also the fact that we have to add an arbitrary function. Can anyone give me a general overview of this method?

\begin{align} \dfrac{\partial^2 \xi}{\partial x\partial y} &= 4x^2y\\ \int\dfrac{\partial^2 \xi}{\partial x\partial y}\mathop{}\!\mathrm{d}x &= 4y\int x^2\mathop{}\!\mathrm{d}x\\ \dfrac{\partial \xi}{\partial y} &= \frac{4x^3y}{3}+\psi(y)\\ \int\dfrac{\partial \xi}{\partial y}\mathop{}\!\mathrm{d}y &= \frac{4x^3}{3}\int y\mathop{}\!\mathrm{d}y+\int\psi(y)\mathop{}\!\mathrm{d}y\\ \xi(x,y) &= \frac{2x^3y^2}{3}+\varphi(x)+\Psi(y)\\ \end{align}

• In the second step, they integrate both sides with respect to $x$. In the fourth, they integrate both sides with respect to $y$. There is no such thing as "partial" integration -- only normal integration is being used here. Mar 8, 2014 at 19:52
• You need to add the arbitrary function of $y$ when integrating with respect to $x$ because many indefinite integrals are possible, and they all differ by a function that is constant in $x$. Here, any function constant with respect to $x$ must be a function in $y$. Mar 8, 2014 at 19:54
• @Potato Why WolframAlpha return "+ constant" instead of a function? wolframalpha.com/input/?i=integrate+4x%5E2y+dx Mar 9, 2014 at 21:56
• 'Constant' means 'constant with respect the variable of integration.' So any function of the other variables works. Mar 10, 2014 at 7:47