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Recently, i had to solve a question in Physics involving partial derivatives:

Find the potential function $V(x,y)$ of an electrostatic field $\vec{E}=2axy\hat i+a(x^2-y^2)\hat j$, where $a$ is a constant.

(A)$V_0+ax^2y-ay^3$

(B)$V_0-ax^2y-ay^3$

(C) $V_0+ax^2y+ay^3$

(D) $\color{green}{V_0-ax^2y+ay^3}$

Note: $E_x=-\dfrac{\partial V}{\partial x}$, $E_y=-\dfrac{\partial V}{\partial y}$

It is easy to obtain the answer by partial differentiation of the options. But how to solve such questions subjectively? What is the method to solve partial differential equations? I am totally new to integration of partial differential equations.

Also, what are such equations and the process of integrating such equations actually called? Because i tried searching multivariable calculus on Wikipedia, but did not find anything useful.

Thanks!

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    $\begingroup$ Here's a good explanation: math.wisc.edu/~conrad/f07/potentials.pdf $\endgroup$ – Hans Lundmark May 4 '16 at 11:00
  • $\begingroup$ @HansLundmark Thank you! That is a very helpful link! I have converted it into an answer. $\endgroup$ – FreezingFire May 4 '16 at 17:19
  • $\begingroup$ That's a good idea! I was too lazy to do it myself... ;-) $\endgroup$ – Hans Lundmark May 4 '16 at 17:20
  • $\begingroup$ @HansLundmark Do you know what this method is called in mathematics? Something like multivariable calculus? Just a supplementary question... $\endgroup$ – FreezingFire May 4 '16 at 17:23
  • $\begingroup$ I don't think it has a standard name. "The usual method for finding the potential of a vector field", perhaps? $\endgroup$ – Hans Lundmark May 4 '16 at 18:00
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@Hans Lundmark's comment solved the question pretty well. I am just transcribing it into an answer over here, in the form of the solution to my question.

To solve the equation $\vec{E}=2axy\,\hat i+a(x^2-y^2)\,\hat j$, we first write: $$ \dfrac{\partial V}{\partial x}=-2axy \\ \therefore V=ax^2y + C(y) $$ where $C(y)$ is a function of $y$. We also have: $$ \dfrac{\partial V}{\partial y}=ax^2-ay^2=ax^2+\dfrac{\mathrm d(C(y))}{\mathrm dy} \\ \therefore C(y)=-\dfrac{ay^3}{3}+C_1 \\ \therefore \bbox[border:1px solid black]{ V=ax^2-\dfrac{ay^3}{3}+C_1} $$ This method can be extended to three variables $x$, $y$ and $z$ also, as is shown in the link. For that case, during the first case, we assume a function $C(y,z)$ instead of just $C(y)$.

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